Test Index

Ellipse

Section: Mathematics

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Question : 15 of 26

Marks: +1, -0

A vertical line passing through the point (h, 0) intersects the ellipse

\frac{x^2}{4} + \frac{y^2}{3} = 1

at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If

\Delta

(h) = area of the triangle PQR,

\Delta_1

\underset{\frac{1}{2} \le h \le 1}{\max}

\Delta

\Delta_2

\underset{\frac{1}{2} \le h \le 1}{\min}

\Delta

\frac{8}{\sqrt{5}} \Delta_1 - 8 \Delta_2

[JEE Adv 2013 P1]

Your Answer:

Solution:

Point of intersection of tangents at P and Q is R (2 sec

\theta

, 0)

Area of

\Delta

PQR =

\frac{1}{2} \cdot 2\sqrt{3} \sin \theta \cdot \left(2 \sec \theta - 2 \cos \theta \right)

⇒

\Delta

=

2\sqrt{3} \cdot \frac{\sin^3 \theta}{\cos \theta}

; where

\cos \theta \in \left[\frac{1}{4}, \frac{1}{2}\right]

Now

\frac{d\Delta}{d\theta} = \frac{2\sqrt{3} \left| \cos \theta \cdot 3 \sin^2 \theta \cos \theta - \sin^3 \theta (-\sin \theta) \right|}{\cos^2 \theta} > 0

As

\theta

increases,

\Delta

increases ⇒ when cos

\,\theta

decreases,

\Delta

increases

∴

\Delta_{\min}

occurs at

\cos \theta = \frac{1}{2}

, Therefore

\Delta_2 = 2\sqrt{3} \cdot \frac{\left(1 - \frac{1}{4}\right)^{3/2}}{\frac{1}{2}} = 4\sqrt{3} \cdot \frac{3\sqrt{3}}{8} = \frac{36}{8}

\Delta_{\max}

occurs at

\cos \theta = \frac{1}{4}

, Therefore

\Delta_1 = 2\sqrt{3} \cdot \frac{\left(1 - \frac{1}{16}\right)^{3/2}}{\frac{1}{4}} =

8\sqrt{3} \cdot \frac{15 \cdot \sqrt{15}}{4 \cdot 4 \cdot 4} = \frac{2\sqrt{3} \cdot 15 \cdot \sqrt{3} \sqrt{5}}{16}

⇒

\Delta_1 = \frac{45}{8} \sqrt{5}

Now

\frac{8}{\sqrt{5}} \Delta_1 - 8 \Delta_2 = 45 - 36 = 9

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