Application of Derivatives 4 Question 60
####63. 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 $Q$. If the tangents to the ellipse at $P$ and $Q$ meet at the point $R$.
If $\Delta(h)=$ area of the $\triangle P Q R, \Delta_{1}=\max {1 / 2 \leq h \leq 1} \Delta(h)$ and $\Delta{2}=\min {1 / 2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}} \Delta{1}-8 \Delta_{2}$ is equal to (2013 Adv)
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Solution:
- PLAN As to maximise or minimise area of triangle, we should find area in terms of parametric coordinates and use second derivative test.
Here, tangent at $P(2 \cos \theta, \sqrt{3} \sin \theta)$ is
$ \begin{aligned} & \frac{x}{2} \cos \theta+\frac{y}{\sqrt{3}} \sin \theta=1 \\ & \therefore \quad R(2 \sec \theta, 0) \\ & \Rightarrow \quad \Delta=\text { Area of } \triangle P Q R \\ & =\frac{1}{2}(2 \sqrt{3} \sin \theta)(2 \sec \theta-2 \cos \theta) \\ & =2 \sqrt{3} \cdot \sin ^{3} \theta / \cos \theta \end{aligned} $
When
$ \frac{1}{4} \leq \cos \theta \leq \frac{1}{2} $
$\therefore \quad \Delta_{1}=\Delta_{\max }$ occurs at $\cos \theta=\frac{1}{4}=\frac{2 \sqrt{3} \cdot \sin ^{3} \theta}{\cos \theta}$
When $\cos \theta=\frac{1}{4}=\frac{45 \sqrt{5}}{8}$
$\Delta_{2}=\Delta_{\min }$ occurs at $\cos \theta=\frac{1}{2}$
$ =\frac{2 \sqrt{3} \sin ^{3} \theta}{\cos \theta} $
When $\quad \cos \theta=\frac{1}{2}=\frac{9}{2}$
$\therefore \quad \frac{8}{\sqrt{5}} \Delta_{1}-8 \Delta_{2}=45-36=9$
