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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