Chapter 11.5 - Simplest Equation of Ellipse

In the previous section, we saw the basic properties of an ellipse. In this section, we will see equation of an ellipse.

• To write the equation of an ellipse, we must first place it on the Cartesian plane.
• The equation will be in the simplest form when the following three conditions are satisfied:
    ♦ The center of the ellipse is at the origin O.
    ♦ The major axis of the ellipse lies along the x-axis.
    ♦ The minor axis of the ellipse lies along the y-axis.
• This is shown in fig.11.34 below:

Fig.11.34

• Based on fig.11.34, we can derive the equation in 6 steps:
1. Let P(x,y) be any point on the ellipse.
2. We know that, F₁ is at a distance of ‘c’ from O.
• So the coordinates of F₁ will be (-c,0)   
3. We know that, F₂ is at a distance of ‘c’ from O.
• So the coordinates of F₂ will be (c,0)
4. Now we have three points and their coordinates:
P(x,y), F₁(-c,0), F₂(c,0)
• Using the distance formula, we can write some distances:

• First we write the distance PF₁:
$\begin{array}{ll} {}&{PF_1} & {~=~}& {\sqrt{(x~-~ -c)^2~+~(y - 0)^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(x + c)^2~+~y^2}} &{} \\ \end{array}$ 

• Next we write the distance PF₂:
$\begin{array}{ll} {}&{PF_2} & {~=~}& {\sqrt{(x~-~ c)^2~+~(y - 0)^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(x - c)^2~+~y^2}} &{} \\ \end{array}$

• Sum of the above two distances is: $\sqrt{(x + c)^2~+~y^2}~+~\sqrt{(x - c)^2~+~y^2}$

5. We know that, B is at a distance of 'a' from O. So the coordinates of B will be (a,0).
• Now we have three points and their coordinates:
B(a,0), F₁(-c,0), F₂(c,0)
• Using the distance formula, we can write some distances:

• First we write the distance BF₁:
$\begin{array}{ll} {}&{BF_1} & {~=~}& {\sqrt{(a~-~ -c)^2~+~(0 - 0)^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(a + c)^2}} &{} \\ {}&{} & {~=~}& {a + c} &{} \\ \end{array}$ 

• Next we write the distance BF₂:
$\begin{array}{ll} {}&{BF_2} & {~=~}& {\sqrt{(a~-~ c)^2~+~(0 - 0)^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(a - c)^2}} &{} \\ {}&{} & {~=~}& {a - c} &{} \\ \end{array}$

• Sum of the above two distances is: (a+c) + (a-c) = 2a

6. Both P and B are points on the same ellipse. So the sum of the distances must be equal.
Equating the results in (4) and (5), we get:

$\begin{array}{ll} {}&{\sqrt{(x + c)^2~+~y^2}~+~\sqrt{(x - c)^2~+~y^2}} & {~=~}& {2a} &{} \\ {\Rightarrow}&{\sqrt{(x + c)^2~+~y^2}} & {~=~}& {2a~-~\sqrt{(x - c)^2~+~y^2}} &{} \\ {\Rightarrow}&{(x + c)^2~+~y^2} & {~=~}& {4a^2~-~4a \sqrt{(x - c)^2~+~y^2}~+~(x - c)^2~+~y^2~~ \color {green} {\text{- - - (I)}}} &{} \\ {\Rightarrow}&{x^2 + 2xc + c^2 + y^2} & {~=~}& {4a^2~-~4a \sqrt{(x - c)^2~+~y^2}~+~x^2 - 2xc + c^2~+~y^2} &{} \\ {\Rightarrow}&{2xc} & {~=~}& {4a^2~-~4a \sqrt{(x - c)^2~+~y^2}- 2xc} &{} \\ {\Rightarrow}&{4xc} & {~=~}& {4a^2~-~4a \sqrt{(x - c)^2~+~y^2}} &{} \\ {\Rightarrow}&{xc} & {~=~}& {a^2~-~a \sqrt{(x - c)^2~+~y^2}~~ \color {green} {\text{- - - (II)}}} &{} \\ {\Rightarrow}&{\frac{xc}{a}} & {~=~}& {a~-~\sqrt{(x - c)^2~+~y^2}} &{} \\ {\Rightarrow}&{\sqrt{(x - c)^2~+~y^2}} & {~=~}& {a~-~\frac{xc}{a}~~ \color {green} {\text{- - - (III)}}} &{} \\ {\Rightarrow}&{(x - c)^2~+~y^2} & {~=~}& {a^2~-~\frac{2axc}{a}~+~\frac{x^2 c^2}{a^2}} &{} \\ {\Rightarrow}&{(x - c)^2~+~y^2} & {~=~}& {a^2~-~2cx~+~\frac{x^2 c^2}{a^2}} &{} \\ {\Rightarrow}&{x^2 - 2cx + c^2~+~y^2} & {~=~}& {a^2~-~2cx~+~\frac{x^2 c^2}{a^2}} &{} \\ {\Rightarrow}&{x^2 + c^2~+~y^2} & {~=~}& {a^2~+~\frac{x^2 c^2}{a^2}} &{} \\ {\Rightarrow}&{x^2 ~+~y^2~-~\frac{x^2 c^2}{a^2}} & {~=~}& {a^2 - c^2} &{} \\ {\Rightarrow}&{x^2 \left(1~-~\frac{c^2}{a^2} \right)~+~y^2} & {~=~}& {a^2 - c^2} &{} \\ {\Rightarrow}&{x^2 \left(\frac{a^2~-~c^2}{a^2} \right)~+~y^2} & {~=~}& {a^2 - c^2~~ \color {green} {\text{- - - (IV)}}} &{} \\ {\Rightarrow}&{x^2 \left(\frac{b^2}{a^2} \right)~+~y^2} & {~=~}& {b^2~~ \color {green} {\text{- - - (V)}}} &{} \\ {\Rightarrow}&{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}} & {~=~}& {1} &{} \\ \end{array}$

◼ Remarks:
• Line marked as (I):
In this line, we square both sides.
• Line marked as (II):
In this line, we divide both sides by 4.
• Line marked as (III):
In this line, we square both sides.
• Line marked as (IV):
In this line, write b² in the place of a² - c².
• Line marked as (V):
In this line, we divide both sides by b².

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Using the above 6 steps, we derived an equation. Now we will prove the converse. It can be written in 9 steps:

1. We derived an equation: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
2. To prove the converse, we assume a point P.
• Let P(x,y) be any point on the ellipse.
• Distance of P from F₁ can be written as:

$\begin{array}{ll} {}&{PF_1} & {~=~}& {\sqrt{(x~-~ -c)^2~+~(y - 0)^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(x + c)^2~+~y^2}} &{} \\ \end{array}$

3. But based on the equation written in (1), we can write:

$\begin{array}{ll} {}&{\frac{y^2}{b^2}} & {~=~}& {1-\frac{x^2}{a^2}} &{} \\ {\Rightarrow}&{y^2} & {~=~}& {b^2\left(1-\frac{x^2}{a^2} \right)} &{} \\ \end{array}$

4. Substituting the above result in (2), we get:

$\begin{array}{ll} {}&{PF_1} & {~=~}& {\sqrt{(x + c)^2~+~y^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(x + c)^2~+~b^2\left(1-\frac{x^2}{a^2} \right)}} &{} \\ {}&{} & {~=~}& {\sqrt{(x + c)^2~+~\left(a^2 - c^2 \right) \left(1-\frac{x^2}{a^2} \right)}~~ \color {green} {\text{- - - (I)}}} &{} \\ {}&{} & {~=~}& {\sqrt{x^2 + 2cx + c^2~+~a^2 - x^2 - c^2 + \frac{c^2 x^2}{a^2}}} &{} \\ {}&{} & {~=~}& {\sqrt{2cx~+~a^2 + \frac{c^2 x^2}{a^2}}} &{} \\ {}&{} & {~=~}& {\sqrt{\left(a+\frac{cx}{a} \right)^2}} &{} \\ {}&{} & {~=~}& {a+\frac{cx}{a}} &{} \\ \end{array}$

◼ Remarks:
• Line marked as (I):
In this line, we write a² - c² in the place of b².

5. Now we consider the distance of P from F₂. It can be written as:

$\begin{array}{ll} {}&{PF_2} & {~=~}& {\sqrt{(x~-~ c)^2~+~(y - 0)^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(x - c)^2~+~y^2}} &{} \\ \end{array}$

6. As we did in the case of PF₁, here also, we substitute for y². We get:

$\begin{array}{ll} {}&{PF_2} & {~=~}& {\sqrt{(x - c)^2~+~y^2}} &{} \\ {}&{} & {~=~}& {\sqrt{(x - c)^2~+~b^2\left(1-\frac{x^2}{a^2} \right)}} &{} \\ {}&{} & {~=~}& {\sqrt{(x - c)^2~+~\left(a^2 - c^2 \right) \left(1-\frac{x^2}{a^2} \right)}~~ \color {green} {\text{- - - (I)}}} &{} \\ {}&{} & {~=~}& {\sqrt{x^2 - 2cx + c^2~+~a^2 - x^2 - c^2 + \frac{c^2 x^2}{a^2}}} &{} \\ {}&{} & {~=~}& {\sqrt{-2cx~+~a^2 + \frac{c^2 x^2}{a^2}}} &{} \\ {}&{} & {~=~}& {\sqrt{\left(a - \frac{cx}{a} \right)^2}} &{} \\ {}&{} & {~=~}& {a - \frac{cx}{a}} &{} \\ \end{array}$

◼ Remarks:
• Line marked as (I):
In this line, we write a² - c² in the place of b². 

7. So the sum of the distances of P from F₁ and F₂ is:
(PF₁ + PF₂) = (a + xc/a) + (a - xc/a) = 2a.

8. Consider step (5) below fig.11.34 at the beginning of this section. We saw that, sum of the distances of point B from the foci is '2a'.

9. So any point P(x,y) on the ellipse will satisfy the equation $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
• The converse is proved.

◼ So we can write:
If the center of the ellipse is at O, major axis lies along the x-axis and minor axis lies along the y-axis, then equation of the ellipse is: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$

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Based on the above equation of the ellipse, we can write two interesting facts:
Fact 1:
This can be written in 5 steps:
1. We have: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
2. This can be rearranged as: $\frac{x^2}{a^2}~=~1~-~\frac{y^2}{b^2}$
• So $\frac{x^2}{a^2}$ will be always less than 1.
• That is: $\frac{x^2}{a^2}~\le~1$
⇒ $x^2 ~\le~a^2$
3. Solving the above inequality, we get:
    ♦ x should not be less than -a.
    ♦ x should not be greater than a.
• That is: $-a~\le~x~\le~a$
4. So we can write:
• Consider any point on the ellipse.
    ♦ The x-coordinate of that point will be greater than -a.
    ♦ The x-coordinate of that point will be less than a.
5. So the ellipse will lie between two vertical lines.
    ♦ The left vertical line is x = -a.
    ♦ The right vertical line is x = a.

Fact 2:
This can be written in 5 steps:
1. We have: $\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1$
2. This can be rearranged as: $\frac{y^2}{b^2}~=~1~-~\frac{x^2}{a^2}$
• So $\frac{y^2}{b^2}$ will be always less than 1.
• That is: $\frac{y^2}{b^2}~\le~1$
⇒ $y^2 ~\le~b^2$
3. Solving the above inequality, we get:
    ♦ y should not be less than -b.
    ♦ y should not be greater than b.
• That is: $-b~\le~y~\le~a$
4. So we can write:
• Consider any point on the ellipse.
    ♦ The y-coordinate of that point will be greater than -b.
    ♦ The y-coordinate of that point will be less than b.
5. So the ellipse will lie between two horizontal lines.
    ♦ The upper horizontal line is y = b.
    ♦ The lower horizontal line is y = -b.

• The ellipse and the lines in fig.11.35 below, demonstrates the two facts:

Fig.11.35

• Equation of the ellipse in the above fig. is: $\frac{x^2}{5^2}~+~\frac{y^2}{3^2}~=~1$
• We see that:
    ♦ Value of 'a' is 5.
        ✰ The vertical lines are related to 5.
    ♦ Value of 'b' is 3.
        ✰ The horizontal lines are related to 3.

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• In this section, we saw a simplest equation of an ellipse.
    ♦ The major axis lies along the x-axis.
    ♦ The minor axis lies along the y-axis.
• We will get another simplest equation also when:
    ♦ The major axis lies along the y-axis.
    ♦ The minor axis lies along the x-axis.
• We will see it in the next section.

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Chapter 11.4 - Ellipse

In the previous section, we completed a discussion on parabola. In this section, we will see ellipse.

Some basics about ellipse can be written in 6 steps:
1. Consider the five points F₁, F₂, P₁, P₂ and P₃ marked in fig.11.29 below:

Fig.11.29

2. Let us write some distances:
• Distance of P₁:
   ♦ Distance of P₁ from F₁ is 9.8 units.
   ♦ Distance of P₁ from F₂ is 21 units.
         ✰ The sum of the two distances is (9.8+21) = 30.8 units.
• Distance of P₂:
   ♦ Distance of P₂ from F₁ is 24.5 units.
   ♦ Distance of P₂ from F₂ is 6.3 units.
         ✰ The sum of the two distances is (24.5+6.3) = 30.8 units.
• Distance of P₃:
   ♦ Distance of P₃ from F₁ is 11.9 units.
   ♦ Distance of P₃ from F₂ is 18.9 units.
         ✰ The sum of the two distances is (11.9+18.9) = 30.8 units.
3. So the three points P₁, P₂ and P₃ have a specialty. It can be written in two steps:
(i) Take any one of those three points.
   ♦ Measure the distance of that point from F₁.
   ♦ Measure the distance of that point from F₂.
(ii) The sum of the two distances will be 30.8 units.
4. There are infinite number of points for which the sum is 30.8 units.      
• All such points will lie in the red curve.
• The red curve is called an ellipse.
5. So we can write the definition:
An ellipse is the set of all points in a plane sum of whose distances from two fixed points in the plane is a constant.
6. Note that, for a particular ellipse, the two points F₁ and F₂ are fixed. If we change one or both of those points, we will get another ellipse.

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Now we will see some basic features of ellipse. They can be written in 7 steps:
1. We have seen that, F₁ and F₂ are two fixed points.
• They are called the foci of the ellipse.
(‘foci’ is the plural of ‘focus’)
2. Draw a line connecting the two foci. Extend this line in both directions.
• Let it intersect the ellipse at A and B.
• Then the line segment AB is called the major axis of the ellipse.
• This is shown in fig.11.30 below:

Fig.11.30

3. The midpoint of F₁F₂ is called center of the ellipse. It is denoted by the letter O.
4. Draw a line through O and perpendicular to AB.
• Let this line intersect the ellipse at C and D
• Then the line segment CD is called the minor axis of the ellipse.
5. The end points A and B of the major axis are called vertices of the ellipse.
6. The length of the major axis is written as ‘2a’. This is shown in fig.11.31 below:

Fig.13.31

• The length of the minor axis is written as ‘2b’
• The distance between the two foci is written as ‘2c’
7. Based on the above fig.11.31, we can write:
   ♦ Length of semi major axis is ‘a’.   
   ♦ Length of semi minor axis is ‘b’.    
   ♦ Distance from center to any one focus is 'c'.

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Now we will derive the relation between 'a', 'b' and 'c'. It can be derived in four steps:
1. In fig.12.32 below, B and C are two points on the ellipse.
2. Let us write the distances related to B:
• Distance of B from F₁ can be written as:

$\begin{array}{ll} {}&{BF_1} & {~=~}& {OF_1} &{~+~}&{OF_2}&{~+~}&{B F_2} \\ {}&{} & {~=~}& {c} &{~+~}&{c}&{~+~}&{OB - OF_2} \\ {}&{} & {~=~}& {c} &{~+~}&{c}&{~+~}&{a - c} \\ {}&{} & {~=~}& {c} &{}&{}&{~+~}&{a} \\ \end{array}$

• Distance of B from F₂ can be written as:

$\begin{array}{ll} {}&{BF_2} & {~=~}& {OB} &{~-~}&{OF_2}&{}&{} \\ {}&{} & {~=~}& {a} &{~-~}&{c}&{}&{} \\ \end{array}$

• So sum of the two distances = (c+a) + (a-c) = 2a

3. Let us write the distances related to C.
• Distance of C from F₁ can be written as:

$\begin{array}{ll} {}&{CF_1} & {~=~}& {\sqrt{(O F_1)^2 ~+~(OC)^2}} &{}&{}&{}&{} \\ {}&{} & {~=~}& {\sqrt{c^2 ~+~b^2}} &{}&{}&{}&{} \\ \end{array}$

• Distance of C from F₂ can be written as:

$\begin{array}{ll} {}&{CF_2} & {~=~}& {\sqrt{(O F_2)^2 ~+~(OC)^2}} &{}&{}&{}&{} \\ {}&{} & {~=~}& {\sqrt{c^2 ~+~b^2}} &{}&{}&{}&{} \\ \end{array}$

• So sum of the two distances
= $\sqrt{c^2 ~+~b^2} + \sqrt{c^2 ~+~b^2}~=~2 \sqrt{c^2 ~+~b^2}$

4. Both B and C are points on the same ellipse. So we can equate the sum of the distances. We get:

$\begin{array}{ll} {}&{2a} & {~=~}& {2\sqrt{c^2 ~+~b^2}} &{}&{}&{}&{} \\ {\Rightarrow}&{a} & {~=~}& {\sqrt{c^2 ~+~b^2}} &{}&{}&{}&{} \\ {\Rightarrow}&{a^2} & {~=~}& {b^2 ~+~c^2} &{}&{}&{}&{} \\ \end{array}$

• This is a simple relation between 'a', 'b' and 'c'.

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The relation a² = b² + c² is applicable to any ellipse. Let us see two cases where this relation gives interesting results.

Case 1: Special case of an ellipse where it becomes a circle.
This can be written in five steps:
1. Suppose that, the vertices A and B are fixed in position.
    ♦ When A and B are fixed, the length 2a becomes a constant.
    ♦ When 2a is a constant, ‘a’ is a constant.
2. Keeping ‘a’ constant, we decrease ‘c’.
• We have the relation: a² = b² + c².
• So, keeping ‘a’ constant, if we decrease ‘c’, The length ‘b’ will automatically increase.
3. An increase in ‘b’ is an increase in the length of minor axis.
• As the length of the minor axis increase, the shape of the ellipse will become more and more circular.
• This is shown in the animation below:

Fig.13.32

4. We are decreasing ‘c’.
• That means, the foci get closer and closer to O.
• Finally, when ‘c’ becomes zero, the foci will merge at the center O.
• In such a situation, ‘a’ and ‘b’ are equal. The ellipse has become a circle.
5. So we can write:
A circle is an ellipse with a = b and c = 0.

Case 2: Special case of an ellipse where it becomes a line.
This can be written in five steps:
1. Suppose that, the vertices A and B are fixed in position.
    ♦ When A and B are fixed, the length 2a becomes a constant.
    ♦ When 2a is a constant, ‘a’ is a constant.
2. Keeping ‘a’ constant, we increase ‘c’.
• We have the relation: a² = b² + c².
• So, keeping ‘a’ constant, if we increase ‘c’, The length ‘b’ will automatically decrease.
3. A decrease in ‘b’ is a decrease in the length of minor axis.
• As the length of the minor axis decrease, the shape of the ellipse will become more and more linear.
• This is shown in the animation below:

Fig.11.33

4. We are increasing ‘c’.
• That means, the foci get closer and closer to vertices A and B.
• Finally, when ‘c’ becomes 'a', the foci will merge with A and B.
• In such a situation, ‘b’ is zero. The ellipse has become a line.
5. So we can write:
A line is an ellipse with a = c and b = 0.

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Eccentricity of an Ellipse

This can be explained in 5 steps:
1. Consider the distance ‘c’.
• It is the distance of the focus from the center O.
2. Consider the distance ‘a’.
• It is the distance of the vertex from the center O.
3. Eccentricity of an ellipse is the ratio of the above two items. It is denoted by the letter ‘e’.
So we can write: $\rm{e=\frac{c}{a}}$
4. Based on this result, we can write: c = ae.
• That means, in any ellipse, the focus is at a distance of ae from the center.
5. We have seen that, for a circle, a = c
• So for a circle, c = ce, which gives e = 1
• We can write:
For a circle, eccentricity is 1

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In the next section, we will see the standard equations of an ellipse.

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Chapter 11.3 - Latus Rectum of Parabola

In the previous section, we saw the basic details about parabolas. In this section, we will see the latus rectum of a parabola.

Latus rectum of a parabola

This can be explained in three steps:
1. Latus rectum is a line segment.
2. If a line segment is to qualify as the latus rectum of a parabola, it must satisfy three conditions.
(i) It must pass through F.
(ii) It must be perpendicular to the axis.
(iii) It’s end points must lie on the parabola.
3. Line segment AB in fig.11.23 below satisfies all three conditions. So it is the latus rectum of that parabola.

Fig.11.23

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Length of the latus rectum

• We have seen the four cases where the equation of parabola is in the simplest form. Length of the latus rectum can be calculated very easily in those four cases.
• Let us see case A. We want the length AB. It can be calculated in 5 steps:
1. In the fig.11.23 above, a perpendicular is drawn from A to the directrix. C is the foot of the perpendicular.
2. Consider the quadrilateral ACDF. We must prove that, ACDF is a rectangle. The proof can be written in 5 steps:
(i) We know that the directrix is perpendicular to axis. So ∠CDF = 90^o
(ii) We know that latus rectum is perpendicular to the axis. So ∠AFD = 90^o
(iii) We have drawn AC perpendicular to the directrix. So ∠ACD = 90^o
(iv) The sum of all interior angles of a quadrilateral is 360^o. Here we have calculated the value of three interior angles. Each of them are 90^o. So the fourth angle ∠CAF must also be 90^o
(v) Since all four interior angles are 90^o, the quadrilateral ACDF is a rectangle.
3. In a rectangle, opposite sides are equal. So AC must be equal to DF
• But DF = (DV + VF) = (a+a) = 2a
• So AC = DF = 2a
4. Point A is on the parabola. It is equidistant from the directrix and F
• So AC = AF
• Thus we get, AF = AC = 2a
5. The parabola is symmetrical about it’s axis. So length BF will be equal to length AF
• So we get: AB = (AF + BF) = (AF + AF) = 2AF = 2 × 2a = 4a

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• Let us see case C. It is shown in fig.11.24 below:

Fig.11.24

• We want the length AB. It can be calculated in 5 steps:
1. In the fig.11.24 above, a perpendicular is drawn from A to the directrix. C is the foot of the perpendicular.
2. Consider the quadrilateral ACDF. We must prove that, ACDF is a rectangle. The proof can be written in 5 steps:
(i) We know that the directrix is perpendicular to axis. So ∠CDF = 90^o
(ii) We know that latus rectum is perpendicular to the axis. So ∠AFD = 90^o
(iii) We have drawn AC perpendicular to the directrix. So ∠ACD = 90^o
(iv) The sum of all interior angles of a quadrilateral is 360^o. Here we have calculated the value of three interior angles. Each of them are 90^o. So the fourth angle ∠CAF must also be 90^o
(v) Since all four interior angles are 90^o, the quadrilateral ACDF is a rectangle.
3. In a rectangle, opposite sides are equal. So AC must be equal to DF
• But DF = (DV + VF) = (a+a) = 2a
• So AC = DF = 2a
4. Point A is on the parabola. It is equidistant from the directrix and F
• So AC = AF
• Thus we get, AF = AC = 2a
5. The parabola is symmetrical about it’s axis. So length BF will be equal to length AF
• So we get: AB = (AF + BF) = (AF + AF) = 2AF = 2 × 2a = 4a

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• In all four cases, we will find that, length of latus rectum is 4a. Where 'a' is the distance of F from V.
• The reader may write the steps for case B and case D.

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Now we will see some solved examples:

Solved example 11.5
For the parabola y² = 8x, write the following items:
(i) Coordinates of the focus
(ii) Equation of the axis
(iii) Equation of the directrix
(iv) Length of the latus rectum.
Solution:
1. Consider the chart that we saw in fig.11.22 of the previous section. It is shown again below:

Fig.11.22

• In our present case, the given equation falls in the category y² = 4ax.
• So we can write:
The axis of the given parabola coincides with the x-axis. And also, the given parabola opens to the right.
2. Comparing y² = 4ax and the given equation y² = 8x, we get:
8 = 4a which gives a = 2
3. Based on the information in the above two steps, we can write:
Focus F lies on the +ve side of the x-axis. It lies at a distance of a = 2 from the origin.
• So the coordinates of F are: (2,0)
• This is the answer for part (i).
4. The axis of the given parabola coincides with the x-axis.
• So equation of the axis of the parabola is: y = 0
• This is the answer for part (ii)
5. The directrix is perpendicular to the x-axis. So it will be parallel to the y-axis.
• The directrix intersects the x-axis at a point a = 2 units away from the origin.
    ♦ This point of intersection will be on the -ve side of the x-axis.
    ♦ So the equation of the directrix will be x = -2.
• This is the answer for part (iii)
6. The length of the latus rectum will be 4a, which gives 4 × 2 = 8 units.
• This is the answer for part (iv)
7. The actual plot is shown below:

Fig.11.25

Solved example 11.6
Find the equation of the parabola with focus (2,0) and directrix x = -2
Solution:
1. Given that, the focus is (2,0).
• So the axis of the parabola passes through (2,0).
• But using this information, we cannot decide about the direction of the axis of the parabola.
2. To help us decide about the direction of the axis, we are given the equation of the directrix. The equation is: x = -2
• Based on this equation, we can write:
Directrix is a vertical line. It passes through (-2,0)
3. If the directrix is a vertical line, the axis of the parabola will be a horizontal line.
• A horizontal line passing through (2,0) is the x-axis itself.
• So we can write:
The axis of the parabola coincides with the x-axis.
• Now we can draw a rough sketch as shown below:

Fig.11.26

• Based on the rough sketch and the chart in fig.11.22, we can write:
The equation of the parabola will be in the form: y² = 4ax
4. So our next aim is to find ‘a’.
• The value of ‘a’ can be calculated in any of the two ways:
(i) ‘a’ is the distance DV, which is 2
(ii) ‘a’ is the distance FV, which is 2
5. So the equation of the parabola is:
y² = 4 × 2 × x
⇒ y² = 8x

Solved example 11.7
Find the equation of the parabola with vertex at (0,0) and focus at (0,2)
Solution:
1. Given that, the focus is (2,0).
• So the axis of the parabola passes through (2,0).
• But using this information, we cannot decide about the direction of the axis.
2. To help us decide about the direction of the axis, we are given the coordinates of the vertex. The coordinates are: (0,0)
• The axis of the parabola is a line which passes through both vertex and focus.
• A line which passes through (0,0) and (2,0) is the x-axis.
• So we can write:
The axis of the parabola coincides with the x-axis.
3. So we have V, F and the axis. We can draw a rough sketch as shown in fig.11.27 below:

Fig.11.27

• Based on the rough sketch and the chart in fig.11.22, we can write:
The equation of the parabola will be in the form: y² = 4ax
4. So our next aim is to find ‘a’.
• ‘a’ is the distance DV, which is 2
5. So the equation of the parabola is:
y² = 4 × 2 × x
⇒ y² = 8x

Solved example 11.8
Find the equation of the parabola which passes through (2,-3) if it is symmetric about the y-axis and it’s vertex is at the origin.
Solution:
1. The given parabola satisfies two conditions:
(i) It is symmetric about one of the coordinate axes.
(ii) It’s vertex is at the origin.
• So this parabola is one of the four simplest forms.
2. This parabola is symmetric about the y-axis.
• So based on the chart in fig.11.22, we can write:
The equation will be one of the two below:
(i) x² = 4ay (opening upwards)
(ii) x² = -4ay (opening downwards)
3. Given that, the parabola passes through (2,-3)
• The point (2,-3) lies in the fourth quadrant. So the parabola opens downwards.
• So we can write:
The equation is in the form x² = -4ay
4. Given that, the parabola passes through (2,-3)
• Substituting these coordinates in the equation obtained in (3), we get:
2² = -4 × a × -3
a = 1/3
5. So equation of the parabola is:

$\begin{array}{ll} {}&{x^2} &{}={}& {-4 \times \left(\frac{1}{3} \right) \times x} &{} \\ {\Rightarrow}&{x^2} &{}={}& {\frac{-4x}{3}} &{} \\ \end{array}$

6. The actual plot is shown below:

Fig.11.28

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Link to a few more solved examples is given below:

Exercise 11.2

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In the next section, we will see latus rectum.

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Chapter 11.2 - Simplest Equation of Parabola

In the previous section, we saw the details about circles. In this section, we will see the details about parabolas.

Some basics about parabola can be written in 9 steps:
1. Consider the point P₁ in fig.11.15 below:

Fig.11.15

• This point P₁ satisfies one condition. It can be written in three steps:
(i) The distance of P₁ from the magenta line is P₁P’₁
(Remember that, the distance of a point from a line should be measured in the perpendicular direction. That is., the line segment P₁P’₁ must be perpendicular to the magenta line.)
(ii) The distance of P₁ from the point F is FP₁
(iii) The condition is that, the above two distances are equal.
• That is., P₁P’₁ = FP₁
    ♦ In the fig., both distances are 11.8 units 
2. Consider the point P₂ in the fig.11.15.
This point P₂ satisfies one condition. It can be written in steps:
(i) The distance of P₂ from the magenta line is P₂P’₂
(ii) The distance of P₂ from the point F is FP₂
(iii) The condition is that, the above two distances are equal.
• That is., P₂P’₂ = FP₂
    ♦ In the fig., both distances are 6.58 units 
3. Consider the point P₃ in the fig.11.15.
This point P₃ satisfies one condition. It can be written in steps:
(i) The distance of P₃ from the magenta line is P₃P’₃
(ii) The distance of P₃ from the point F is FP₃
(iii) The condition is that, the above two distances are equal.
• That is., P₃P’₃ = FP₃
    ♦ In the fig., both distances are 16.19 units 
4. We have seen three points. All of them satisfy the same condition:
    ♦ Distance from the magenta line
    ♦ is equal to
    ♦ Distance from F
5. There are infinite number of points which will satisfy the above condition. All such points will lie on the red curve.
• The red curve is called Parabola.
◼ We can write:
• A parabola is the set of all points that are equidistant from a fixed line and a fixed point.
    ♦ The fixed point must not be on the fixed line.
    ♦ The fixed line, fixed point and all points on the parabola, must lie on the same plane.
        ✰ The fixed line is called the directrix (l) of the parabola.     
        ✰ The fixed point is called the focus (F) of the parabola.
6. We can draw a line in such a way that:
    ♦ The line passes through the focus F
    ♦ The line is perpendicular to the directrix.
• Only one line is possible which satisfies both the conditions. That line is called the axis of the parabola.
• It is shown in fig.11.16 below:

Fig.11.16

7. Consider the point where the axis intersects the parabola. That point is called the vertex (V) of the parabola. It is shown in fig.11.16 above.
8. Consider the point where the axis intersects the directrix. That point is marked as D in the above fig.11.16.
• Let us see the significance of point D. It can be written in four steps:
(i) We know that, all points on the parabola are equidistant from l and F
(ii) V is a point on the parabola. So V is also equidistant from l and F
(iii) That means, V is the midpoint of DF
(iv) In other words:
    ♦ Distance of l  from V
    ♦ is equal to
    ♦ Distance of F from V
9. Let us see how the word parabola is derived:
    ♦ The word ‘para’ means ‘for’
    ♦ The word ‘bola’ means ‘throwing’
• In projectile motion in physics classes, we have seen that, a thrown object follows a parabolic path.

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Standard equations of parabola

• The equation of a parabola will be in the simplest form when two conditions are satisfied:
(i) V is at the origin of the coordinate axes.
(ii) Axis of the parabola coincides with one of the coordinate axes.
• Based on this information, we can think of four cases:
A. V is at the origin, axis coincides with the x-axis and parabola opens to the right.
    ♦ This is shown in fig.11.17(a) below.
        ✰ Axes are shown in cyan color
        ✰ Parabola is shown in red color.
        ✰ Directrix is shown in magenta color. It is marked as l.
        ✰ Axis of the parabola is shown in yellow color.
B. V is at the origin, axis coincides with the x-axis and parabola opens to the left.
    ♦ This is shown in fig.11.17(b) below.
C. V is at the origin, axis coincides with the y-axis and parabola opens upwards.
    ♦ This is shown in fig.11.17(c) below.
D. V is at the origin, axis coincides with the y-axis and parabola opens downwards.
    ♦ This is shown in fig.11.17(d) below.

Fig.11.17

• We will now derive equation in each case.

Case A:
V is at the origin, axis coincides with the x-axis and parabola opens to the right.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.18 below:

Fig.11.18

• Remember the two points:
(i) Axis of the parabola coincides with the x-axis. So F lies on the x-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (a,0)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (-a,0)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘-a’
• P₁ and P lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘y’
• Thus the coordinates of P₁ are: (-a,y)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll} {}&{PP_1} &{}={}& {\sqrt{(-a - x)^2~+~(y-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{[-1 × (a + x)]^2~+~(y-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{(a + x)^2}} &{} \\ \end{array}$

• Next we calculate PF:
$\begin{array}{ll} {}&{PF} &{}={}& {\sqrt{(a - x)^2~+~(0-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{(a - x)^2~+~y^2}} &{} \\ \end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll} {}&{PP_1} & {~=~}& {PF} &{}&{}&{} \\ {\Rightarrow}&{\sqrt{(a + x)^2}} & {~=~}& {\sqrt{(a - x)^2~+~y^2}} &{}&{}&{} \\ {\Rightarrow}&{(a + x)^2} & {~=~}& {(a - x)^2~+~y^2} &{}&{}&{} \\ {\Rightarrow}&{a^2 + 2ax + x^2} & {~=~}& {a^2 - 2ax + x^2~+~y^2} &{}&{}&{} \\ {\Rightarrow}&{2ax} & {~=~}& {- 2ax + y^2} &{}&{}&{} \\ {\Rightarrow}&{y^2} & {~=~}& {4ax} &{}&{}&{} \\ \end{array}$

• This is the equation of the parabola for case A.

Case B:
V is at the origin, axis coincides with the x-axis and parabola opens to the left.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.19 below:

Fig.13.19

• Remember the two points:
(i) Axis of the parabola coincides with the x-axis. So F lies on the x-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (-a,0)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (a,0)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘a’
• P₁ and P lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘y’
• Thus the coordinates of P₁ are: (a,y)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll} {}&{PP_1} &{}={}& {\sqrt{(a - x)^2~+~(y-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{(a - x)^2}} &{} \\ \end{array}$

• Next we calculate PF:
$\begin{array}{ll} {}&{PF} &{}={}& {\sqrt{(-a - x)^2~+~(0-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{[-1 × (a + x)]^2~+~y^2}} &{} \\ {}&{} &{}={}& {\sqrt{(a + x)^2~+~y^2}} &{} \\ \end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll} {}&{PP_1} & {~=~}& {PF} &{}&{}&{} \\ {\Rightarrow}&{\sqrt{(a - x)^2}} & {~=~}& {\sqrt{(a + x)^2~+~y^2}} &{}&{}&{} \\ {\Rightarrow}&{a^2 - 2ax + x^2} & {~=~}& {a^2 + 2ax + x^2~+~y^2} &{}&{}&{} \\ {\Rightarrow}&{-2ax} & {~=~}& {2ax + y^2} &{}&{}&{} \\ {\Rightarrow}&{y^2} & {~=~}& {-4ax} &{}&{}&{} \\ \end{array}$

• This is the equation of the parabola for case B.

Case C:
V is at the origin, axis coincides with the y-axis and parabola opens upwards.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.20 below:

Fig.11.20

• Remember the two points:
(i) Axis of the parabola coincides with the y-axis. So F lies on the y-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (0,a)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (0,-a)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘-a’
• P₁ and P lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘x’
• Thus the coordinates of P₁ are: (x,-a)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll} {}&{PP_1} &{}={}& {\sqrt{(x - x)^2~+~(-a-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{[-1 × (a + y)]^2}} &{} \\ {}&{} &{}={}& {\sqrt{(a + y)^2}} &{} \\ \end{array}$

• Next we calculate PF:
$\begin{array}{ll} {}&{PF} &{}={}& {\sqrt{(0 - x)^2~+~(a-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{x^2~+~(a-y)^2}} &{} \\ \end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll} {}&{PP_1} & {~=~}& {PF} &{}&{}&{} \\ {\Rightarrow}&{\sqrt{(a + y)^2}} & {~=~}& {\sqrt{x^2~+~(a-y)^2}} &{}&{}&{} \\ {\Rightarrow}&{a^2 + 2ay + y^2} & {~=~}& {x^2 ~+~ a^2 - 2ay + y^2} &{}&{}&{} \\ {\Rightarrow}&{2ay} & {~=~}& {x^2 - 2ay} &{}&{}&{} \\ {\Rightarrow}&{x^2} & {~=~}& {4ay} &{}&{}&{} \\ \end{array}$

• This is the equation of the parabola for case C.

Case D:
V is at the origin, axis coincides with the y-axis and parabola opens downwards.
• This can be written in 6 steps:
1. Let the focus F be at a distance of ‘a’ units from the vertex V.
• This is shown in fig.11.21 below:

Fig.11.21

• Remember the two points:
(i) Axis of the parabola coincides with the y-axis. So F lies on the y-axis.
(ii) V is at the origin O
• Based on these two points, we can write:
Coordinates of F are (0,-a)
2. Remember that, D and F are equidistant from V.
• So the coordinates of D will be (0,a)
3. Mark any convenient point P(x,y) on the parabola.
• Draw a perpendicular from P onto the directrix l
• Let the foot of the perpendicular be P₁
4. Let us write the coordinates of P₁:
• P₁ and D lies on the same horizontal line.
    ♦ So both will have the same y-coordinate ‘a’
• P₁ and P lies on the same vertical line.
    ♦ So both will have the same x-coordinate ‘x’
• Thus the coordinates of P₁ are: (x,a)
5. Now we have the coordinates of all the points. We can calculate the distances:

• First we calculate PP₁:
$\begin{array}{ll} {}&{PP_1} &{}={}& {\sqrt{(x - x)^2~+~(a-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{(a - y)^2}} &{} \\ \end{array}$

• Next we calculate PF:
$\begin{array}{ll} {}&{PF} &{}={}& {\sqrt{(0 - x)^2~+~(-a-y)^2}} &{} \\ {}&{} &{}={}& {\sqrt{x^2~+~[-1 × (a+y)]^2}} &{} \\ {}&{} &{}={}& {\sqrt{x^2~+~(a+y)^2}} &{} \\ \end{array}$

6. Remember that, any point on the parabola is equidistant from F and l.
• So we can equate the above two distances. We get:
$\begin{array}{ll} {}&{PP_1} & {~=~}& {PF} &{}&{}&{} \\ {\Rightarrow}&{\sqrt{(a - y)^2}} & {~=~}& {\sqrt{x^2~+~(a+y)^2}} &{}&{}&{} \\ {\Rightarrow}&{a^2 - 2ay + y^2} & {~=~}& {x^2 ~+~ a^2 + 2ay + y^2} &{}&{}&{} \\ {\Rightarrow}&{-2ay} & {~=~}& {x^2 + 2ay} &{}&{}&{} \\ {\Rightarrow}&{x^2} & {~=~}& {-4ay} &{}&{}&{} \\ \end{array}$

• This is the equation of the parabola for case D.

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So we have seen all the four cases. The results follow a pattern. So we can draw a flow chart:

Fig.11.22

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In the next section, we will see latus rectum.

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Chapter 11.1 - General Equation of A Circle

In the previous section, we saw the basic details about conic sections. In this section, we will see some advanced details about circles.

A circle can be defined in 5 steps:
1. In fig.11.11(a) below, a curve is drawn in red color.
• A point C is marked inside the curve.

Fig.11.11

2. Mark a few convenient points P₁, P₂, P₃, P₄, P₅, . . .  on the curve.
• In the fig.(a), three points P₁, P₂ and P₃ are marked.
3. Measure the distances CP₁, CP₂, CP₃, CP₄, CP₅, . . .  
• If all those distances are the same, then the curve is a circle.
4. The point C is called the centre of the circle.
5. The distance from C to any point on the circle is called the radius of the circle.
• It is denoted using the letter r.

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Our next task is to derive the equation of a circle. It can be done in 4 steps:
1. In fig.11.12(b) above, the coordinates of C are (h,k)
2. Mark any convenient point P on the circle. Let the coordinates of P be (x,y)
3. Then the distance CP can be calculated using the distance formula. We get:
CP = $\sqrt{(x-h)^2 + (y-k)^2}$ 
4. But CP = r. So we can write:
$r = \sqrt{(x-h)^2 + (y-k)^2}$
• Squaring both sides, we get:
$r^2 = (x-h)^2 + (y-k)^2$
• This can be rearranged as:
$(x-h)^2 + (y-k)^2 = r^2$
• This is the equation of a circle.

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The equation of a circle can be written in expanded form also. It's details can be written in 5 steps:
1. Opening the brackets in the above equation, we get:
r² = x² - 2hx + h² + y² - 2ky + k² 
2. This can be rearranged as:
x² - 2hx + h² + y² - 2ky + k² - r² = 0
3. Writing similar terms together, we get:
x² + y² - 2hx - 2ky + h² + k² - r² = 0
4. We can describe the terms as follows:
• First term is a term in x².
• Second term is a term in y².
• Third term is a term in x.
• Fourth term is a term in y.
• The remaining terms are constant terms.
5. So we can write the equation as:
[x²] + [y²] + [-2hx] + [-2ky] + [h² + k² - r²] = 0
• Based on this we can write:
(i) The coefficient of x² term is equal to 1.
(ii) The coefficient of y² term is equal to 1.
(iii) The coefficient of x term is equal to -2h.
   ♦ So by dividing this coefficient by -2, we will get h.
   ♦ Thus we get the x-coordinate of the centre C.
(iv) The coefficient of y term is equal to -2k.
   ♦ So by dividing this coefficient by -2, we will get k.
   ♦ Thus we get the y-coordinate of the centre C.
(v) The constant term is h² + k² - r².
   ♦ Once we know k, h and the constant term, we can easily calculate the radius r.

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Now we will see some solved examples:
Solved example 11.1
Derive the equation of a circle whose center is at O
Solution:
1. In fig.11.11(c) above, the center of the circle is at O
2. Let P(x,y) be any point on the circle.
3. Using the distance formula, we can write:
OP = $\sqrt{(x-0)^2 + (y-0)^2}~=~\sqrt{x^2 + y^2}$ 
4. But OP = r. So we can write:
$r = \sqrt{x^2 + y^2}$
• Squaring both sides, we get:
$r^2 = x^2 + y^2$
• This can be rearranged as:
$x^2 + y^2 = r^2$
• This is the equation of a circle with radius r and center O.

Easier method:
1. The general equation of a circle with radius r and center at (h,k) is:
$(x-h)^2 + (y-k)^2 = r^2$
2. If the center is at O, both h and k are zero.
• Substituting these in the general equation, we get:
$(x-0)^2 + (y-0)^2 = r^2$   
• This is same as: $x^2 + y^2 = r^2$

Solved example 11.2
Find the equation of the circle with center at (3/2, 7/5) and radius 5 units.
Solution:
1. The general equation of a circle with radius r and center at (h,k) is:
$(x-h)^2 + (y-k)^2 = r^2$
2. The center is given as (3/2, 7/5).
• Substituting these in the general equation, we get:

$\begin{array}{ll} {}&{\left(x-\frac{3}{2}\right)^2 + \left(y-\frac{7}{5}\right)^2} & {~=~}& {6^2} &{}&{}&{} \\ {\Rightarrow}&{x^2 - 3x + \frac{9}{4}~+~y^2 - \frac{14y}{5} + \frac{49}{25}} & {~=~}& {36} &{}&{}&{} \\ {\Rightarrow}&{100 \times \left[x^2 - 3x + \frac{9}{4}~+~y^2 - \frac{14y}{5} + \frac{49}{25} \right]} & {~=~36 \times 100}& {} &{\color{green}{\text{LCM of 4, 5, 25 is 100}}}&{}&{} \\ {\Rightarrow}&{100 x^2 - 300 x + 25 × 9 + 100 y^2 - 20 × 14 × y + 4 × 49} & {~=~3600}& {} &{}&{}&{} \\ {\Rightarrow}&{100 x^2 + 100 y^2 - 300 x - 280 y + 225 + 196} & {~=~3600}& {} &{}&{}&{} \\ {\Rightarrow}&{100 x^2 + 100 y^2 - 300 x - 280 y - 3179} & {~=~0}& {} &{}&{}&{} \\ \end{array}$

3. The actual plot is shown in fig.11.12 below:

Fig.11.12

Solved example 11.3
Find the center and radius of the circle x² + y² + 8x +10y - 8 = 0
Solution:
1. The general equation of a circle is:
[x²] + [y²] + [-2hx] + [-2ky] + [h² + k² - r²] = 0
• The given equation can be written in the general form:
[x²] + [y²] + [8x] + [10y] + [-8] = 0
2. So we can write:
(i) coefficient of x = -2h = 8. So h = -4
(ii) coefficient of y = -2k = 10. So k = -5
(iii) constant term = h² + k² - r² = -8
• Substituting the values of h and k, we get:

$\begin{array}{ll} {}&{(-4)^2 + (-5)^2 - r^2} & {~=~}& {-8} &{}&{}&{} \\ {\Rightarrow}&{16 + 25 - r^2} & {~=~}& {-8} &{}&{}&{} \\ {\Rightarrow}&{41 - r^2} & {~=~}& {-8} &{}&{}&{} \\ {\Rightarrow}&{r^2} & {~=~}& {49} &{}&{}&{} \\ {\Rightarrow}&{r} & {~=~}& {\sqrt{49}} &{}&{}&{} \\ {\Rightarrow}&{r} & {~=~}& {\pm 7} &{}&{}&{} \\ \end{array}$

• Radius cannot be -ve. We must take the +ve value.
3. We can write:
• Center (h,k) of the circle is (-4, -5)
• Radius r of the circle is 7 units.
• Once we get the center and radius, we can use the simple form also:

$\begin{array}{ll} {}&{(x-h)^2 + (y-k)^2} & {~=~}& {r^2} &{}&{}&{} \\ {\Rightarrow}&{(x~-~-4)^2 + (y~-~-5)^2} & {~=~}& {7^2} &{}&{}&{} \\ {\Rightarrow}&{(x+4)^2 + (y+5)^2} & {~=~}& {49} &{}&{}&{} \\ \end{array}$

4. The actual plot is shown below:

Fig.11.13

Solved example 11.4
Find the equation of the circle which passes through the points (2,-2) and (3,4) and whose center lies on the line x+y=2.
Solution:
1.The general equation of a circle is: (x-h)² + (y-k)² = r²
• Where (h,k) is the center and r is the radius.
2. In our present case, the circle passes through (2,-2). Substituting these values in the general equation, we get:
(2-h)² + (-2-k)² = r²
⇒ (2-h)² + [-1(2+k)]² = r²
⇒ (2-h)² + (2+k)² = r²
• Expanding this and rearranging, we get:
8 – 4h + 4k +h² + k² - r² = 0
3. In our present case, the circle passes through (3,4) also. Substituting these values in the general equation, we get:
(3-h)² + (4-k)² = r²
• Expanding this and rearranging, we get:
25 – 6h - 8k + h² + k² - r² = 0
4. So now we have two equations:
(i) From (2), we have: 8 – 4h + 4k +h² + k² - r² = 0
(ii) From (3), we have: 25 – 6h - 8k + h² + k² - r² = 0
5. Subtracting 4(ii) from 4(i), we get: -17 + 2h +12k = 0
6. Given that, the center lies on the line x+y=2.
• We assumed that, the center is (h,k). Substituting these in the equation of the line, we get: h+k=2
• So h = 2-k
7. Substituting this value of h in (5), we get:
-17 + 2(2-k) + 12k = 0
⇒ -17 + 4 – 2k  + 12k = 0
⇒ -13 + 10k = 0
⇒ k = 1.3
8. Substituting this value of k in (6), we get: h = 2-1.3 = 0.7
9. Now we have the center of the circle (h,k) = (0.7, 1.3)
• (2,-2) is a point on the circle.
• So using the distance formula, we can write:
r² = [(2 - 0.7)² + (-2 - 1.3)²] = [1.3² + 3.3²] = 12.58
10. Now we can write the equation of the circle:
(x - 0.7)² + (y - 1.3)² = 12.58
• The actual plot is shown below:

Fig.11.14

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The link below gives a few more solved examples:

Exercise 11.1

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In the next section, we will see parabola.

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Chapter 11 - Conic Sections

In the previous section, we completed a discussion on straight lines. In this chapter, we will see the details about curves like circles, ellipses, parabolas and hyperbolas.

First we will see a double napped right circular cone. It can be written in 5 steps:
1. In fig.11.1(a) below, l is a fixed vertical line.
• m is another line which intersects l at V.
• Also, m is inclined at an angle of 𝛼 with l.

Fig.11.1

2. Next step is to rotate m around l
• During the rotation,
    ♦ Point V must not change.
    ♦ The angle 𝛼 must not change.
• Such a rotation is shown in the animation in fig.11.2 below:

Fig.11.2

We see that:
    ♦ The rotating line m generates a conical surface.
    ♦ The top end of m moves along the top yellow circle.
    ♦ The bottom end of m moves along the bottom yellow circle.
3. We get an upper nappe and a lower nappe. The yellow circles form the bases of the nappes.
• This is shown in fig.11.1(b) above. The shape in fig.11.1(b) is called a double-napped right circular cone. For our discussions in this chapter, we will call it simply as cone.
• The cone obtained in this way will be hollow.
4. Let us see the various features of a cone:
(i) The point V is called vertex of the cone.
(ii) The line l is called axis of the cone.
(iii) The line m is called generator of the cone
(iv) The vertex separates the cone into two parts called nappes.
• These features are marked in fig.11.1(c) above.
5. Keeping V and 𝛼 fixed, we can increase the lengths of l and m.
• Then the size of the cone will also increase.
• By increasing the lengths of l and m, the size of cone can be increased upto infinity.

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Sections of a cone

This can be explained in 5 steps:
1. We can cut a cone using a plane. An example is shown in fig.11.3(a) below:

Fig.11.3

2. When such a cut is made, the cone is separated into two parts.
(i) A larger part. We will call it major part.
(ii) A smaller part. We will call it minor part.
3. After making the cut, we remove the minor part and the plane.
• After removing them, when we look at the portion where the cut is made, we will see a curve.
• This curve is called a conic section. This curve is highlighted in green color in fig.11.3(b).
◼ We can write:
Conic sections are curves obtained by intersecting a right circular cone by a plane.
4. Consider the situation in fig.11.3(a). In this situation, if we look from the edge of the plane, that edge will appear as a line. This is shown in fig.11.3(c).
• The upper and lower nappes will appear as triangles.
• The angle between the plane and the axis l is marked as 𝛽.
5. There are three ways to cut a cone.
(i) The plane can pass through the vertex.
(ii) The plane can pass through the upper nappe.
(iii) The plane can pass through the lower nappe.
• The angle 𝛽 can also vary according to the requirement.

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Circle, ellipse, parabola and hyperbola as conic sections

This can be explained in 6 steps:
1. In the above fig.13.3(c), the angle 𝛽 is less than 90^o.
• If 𝛽 is exactly 90^o, then the plane will cut the cone in a horizontal manner. This is shown in fig.11.4(a) below:

Fig.11.4

• In fig.11.4(b) above, we can clearly see the major part and minor part.
• In fig.11.4(c) above, the plane and the minor part are removed to reveal the conic section.
• The conic section is highlighted in green color. It is a circle.
2. We know the significance of angle 𝛼. It is the angle between the generator m and the axis l.
• If this 𝛼 is 90^o, we will not get a cone. We will get only a plane surface.
• So 𝛼 must be always less than 90^o. We can write: 𝛼 < 90^o
3. We have seen the situation where 𝛽 is 90^o. We saw that, a circle will be obtained.
• Now we will see the situation when 𝛽 is less than 90^o. For this situation, we can write: 𝛼 < 90^o and 𝛽 < 90^o
• Here three cases can arise:
(i)  𝛼 < 90^o, 𝛽 < 90^o and 𝛼 < 𝛽
(ii)  𝛼 < 90^o, 𝛽 < 90^o and 𝛼 = 𝛽
(iii)  𝛼 < 90^o, 𝛽 < 90^o and 𝛼 > 𝛽
4. An ellipse is obtained in case 3(i). It is shown in fig.11.5 below:

Fig.11.5

• In fig.11.5(a) above, 𝛽 is greater than 𝛼.
• In fig.(b), we can clearly see the major and minor parts.
• In fig.(c), the plane and the minor part are removed to reveal the conic section.
• The conic section is highlighted in green color. It is an ellipse.
5. A parabola is obtained in case 3(ii). It is shown in fig.11.6 below:

Fig.11.6

• In fig.11.6(a) above, 𝛽 is equal to 𝛼.
• In fig.(b), we can clearly see the major and minor parts.
• In fig.(c), the plane and the minor part are removed to reveal the conic section.
• The conic section is highlighted in green color. It is a parabola.
6. A hyperbola is obtained in case 3(iii). It is shown in fig.11.7 below:

Fig.11.7

• In fig.11.7(a) above, 𝛽 is less than 𝛼.
    ♦ So the plane is able to cut both upper nappe and lower nappe.
• In fig.(b), we can clearly see the major and minor parts.
• In fig.(c), the plane and the minor parts are removed to reveal the conic section.
• The conic section is highlighted in green color. It is a hyperbola.
    ♦ We see that, a hyperbola has two curves.

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Degenerated conic sections

This can be explained in 4 steps:
1. Degenerated conic sections are special cases when the cutting plane passes through the vertex V.
2. We have seen that 𝛼 must be less than 90^o. [step (2) below fig.11.4]
3. We have also seen that:
𝛽 can be equal to 90^o [fig.11.4]
• In this situation, if the plane passes through the vertex, we can represent it as shown in fig.11.8(a) below:

Fig.11.8

• The section obtained will be a point.
4. We have also seen the cases where 𝛽 is less than 90^o:
(i) 𝛽 can be less than 90^o and greater than 𝛼 [fig.11.5]
• In this situation, if the plane passes through the vertex, we can represent it as shown in fig.11.8(b) above.
• The section obtained will be a point.
(ii) 𝛽 can be less than 90^o and equal to 𝛼 [fig.11.6]
• In this situation, if the plane passes through the vertex, we can represent it as shown in fig.11.8(c) above.
• The plane just touches the lateral surface of the nappes. It does not make a cut.
• The section obtained will be a straight line.
• It is the degenerated case of a parabola.
(iii) 𝛽 can be less than 90^o and less than 𝛼 [fig.11.7]
• In this situation, if the plane passes through the vertex, we can represent it as shown in fig.11.9(a) below:

Fig.11.9

• In fig.11.9(a) above, 𝛽 is less than 𝛼.
    ♦ So the plane is able to cut both upper nappe and lower nappe.
• In fig.(b), we can clearly see the major and minor parts.
• In fig.(c), the plane and the minor parts are removed to reveal the conic section.
• The conic section is highlighted in green color. It is a pair of two intersecting straight lines.
• It is the degenerated case of a hyperbola.
(iv) 𝛽 can be zero (which is less than 90^o) and less than 𝛼
• In this situation, if the plane passes through the vertex, we can represent it as shown in fig.11.10(a) below:

Fig.11.10

• In fig.11.10(a) above, 𝛽 is zero. The axis lies in the plane.
    ♦ The plane is able to cut both upper nappe and lower nappe.
• In fig.(b), we can clearly see the two parts. They cannot be called as major and minor parts. Because, both are of the same size.
• In fig.(c), the plane and one of the parts are removed to reveal the conic section.
• The conic section is highlighted in green color. It is a pair of two intersecting straight lines.
• It is the degenerated case of a hyperbola.

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So we have seen circle, ellipse, parabola and hyperbola. In the next section, we will see more details about circles.

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