In the previous section, we completed a discussion on hyperbola. In this section, we will see some miscellaneous examples.
Solved example 11.18
The focus of the parabolic mirror shown in fig.11.55 below is at a distance of 5 cm from it's vertex. If the mirror is 45 cm deep, find the distance AB.
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| Fig.11.55 |
Solution:
1. From fig.11.55, it is clear that:
♦ Vertex of the mirror is at the origin O.
♦ Axis of the mirror lies along the x-axis.
♦ The parabola opens to the right.
2. So we can draw a rough sketch of the graph of the parabola as shown in fig.11.56 below:
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| Fig.11.56 |
3. General equation of the parabola in the above fig.11.56 is: y2 = 4ax
• Given that, F is at a distance of 5 cm from the vertex. So we get: a = 5 cm.
• So the equation of the parabola is: y2 = 4 × 5 × x = 20x
4. In fig.11.55, 'A' is a point on the parabola. It is at a horizontal distance of 45 cm from the y=axis.
• So the coordinates of 'A' can be written as (45,y)
5. 'B' is also at a distance of 45 cm from the y-axis.
• So 'A' and 'B' are symmetrical points. The coordinates of B can be written as (45,-y)
6. Since A(45,y) is a point on the parabola, we can substitute those coordinates in the equation of the parabola obtained in (3).
• We get: y2 = 20 × 45 = 900
• From this we get: y = 30 or -30
7. So the coordinates of point A are (45,30)
• Also, the coordinates of B are (45,-30)
8. Using the coordinates, we get:
Distance AB = $\sqrt{(45 - 45)^2 + (-30 - 30)^2}~=~\sqrt{60^2} = 60$ cm.
Solved example 11.19
The distance between the supports of a beam is 12 metres. When loads are applied on the beam, it deflects and becomes parabolic in shape. The maximum deflection of 3 cm occurs at the midpoint between the supports. At what point between the supports, does a deflection of 1 cm occur?
Solution:
1. The situation before deflection is shown in fig.11.57(a) below:
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| Fig.11.57 |
• The beam is represented by the red horizontal line AB.
2. After deflection, the the midpoint O of the beam is at a vertical distance of 3 cm below the horizontal line. This is shown in fig.11.57(b).
• Let C be the point where the deflection is 1 cm. Then C will be at a vertical distance of 2 cm above O
3. Given that, the deflected beam is in the shape of a parabola.
• We can consider the lowest point ‘O’ as the vertex.
4. So we can draw a rough sketch of the graph of the parabola as shown in fig.11.58 below:
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| Fig.11.58 |
• Vertex of the parabola is at the origin O. This vertex is the midpoint of the beam.
• A is at a distance of 6 m to the left of midpoint. Also, A is 3 cm (0.03 m) vertically above O.
♦ So the coordinates of A are: (-6,0.03)
• B is at a distance of 6 m to the right of midpoint. Also, B is 3 cm (0.03 m) vertically above O.
♦ So the coordinates of B are: (6,0.03)
• C is at an unknown distance. It is represented by ‘x’. Also, C is 2 cm (0.02 m) vertically above O.
♦ So the coordinates of C are (x,0.02)
5. The parabola in fig.11.58 is of the form x2 = 4ay
6. A(-6,0.03) is a point on the parabola. So substituting the coordinates in (5), we get:
(-6)2 = 4 × a × 0.03
⇒ 36 = 0.12a
⇒ a = 300
7. So the equation of the parabola is: x2 = 4 × 300y = 1200y
8. Substituting the coordinates of C in the above equation, we get:
x2 = 1200 × 0.02 = 24
⇒ x = √24 = 2√6
9. So we can write:
Deflection of 1 cm, occurs at a distance of 2√6 m from the midpoint of the beam.
Solved example 11.20
A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x,y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse.
Solution:
1. In fig.11.59 below, the rod AB is resting between OX and OY.
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| Fig.11.59 |
• AB makes an angle 𝜃 with OX.
• End A is on OX. End B is on OY.
• P is a point on the rod.
♦ Distance of P from A is 6 cm.
♦ Distance of P from B is 9 cm.
2. Horizontal and vertical dashed lines:
• A horizontal dashed line is drawn through P. This line intersects the y-axis at Q.
• A vertical dashed line is drawn through P. This line intersects the x-axis at R.
3. Since OX and PQ are parallel, ∠ QPB = 𝜃
4. Let the coordinates of P be (x,y)
♦ Then PQ wil be equal to x.
♦ Also PR will be equal to y.
5. Now we take trigonometric ratios:
(i) Consider ⧍PBQ.
• In this triangle, cos 𝜃 = x/9.
(ii) Consider ⧍PAR.
• In this triangle, sin 𝜃 = y/6
6. We have the identity: cos2𝜃 + sin2𝜃 = 1
• Substituting the values from (5), we get:
$\begin{array}{ll}
{}&{\left(\frac{x}{9} \right)^2 + \left(\frac{x}{9} \right)^2}
& {~=~}& {1}
&{} \\
{\Rightarrow}&{\frac{x^2}{81} + \frac{y^2}{36}}
& {~=~}& {1}
&{} \\
\end{array}$
7. This is the equation of an ellipse. Any values of (x,y) which the point P takes, will fall on the ellipse.
• So we can write:
Locus of point P is the ellipse $\frac{x^2}{81} + \frac{y^2}{36} = 1$.
◼ More details can be written in 3 steps:
1. The locus is the path traced by a moving point.
• The movement of the point must satisfy one or more specified conditions.
2. In our present problem, P is the point which moves.
The conditions are:
(i) Point A must always be on the x-axis.
(ii) Point B must always be on the y-axis.
(iii) Point P must always be:
♦ on the line connecting A and B..
♦ 6 cm away from A.
♦ 9 cm away from B.
3. All conditions specified in (2) are satisfied in the animation in fig.11.60 below:
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| Fig.11.60 |
The path traced by P is the ellipse $\frac{x^2}{81} + \frac{y^2}{36} = 1$
Link to a few more solved examples is given below:
Miscellaneous ExerciseIn the next chapter, we will see Three dimensional geometry.
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