QJEEAdv13Ma368e: JEE Advanced 2013 Mathematics - Class 11 JEEAdv13Ma368e Solution: A vertical line passing through the point intersects the…

A vertical line passing through $(h, 0)$ is $x=h$ and this vertical line intersect the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at P and Q.

Put $x=h$ in the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$

$\begin{aligned}& \Rightarrow \frac{h^2}{4}+\frac{y^2}{3}=1 \\& \Rightarrow y= \pm \sqrt{3-\frac{3 h^2}{4}}\end{aligned}$

So $\mathrm{P}=\left(h, \sqrt{3-\frac{3 h^2}{4}}\right)$ and $\mathrm{Q}=\left(h,-\sqrt{3-\frac{3 h^2}{4}}\right)$

Equation of tangent of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at P is

$\frac{h x}{4}+\frac{y}{3} \sqrt{3-\frac{3 h^2}{4}}=1 \quad \text{... (i)}$

Equation of tangent of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at Q is

$\frac{h x}{4}-\frac{y}{3} \sqrt{3-\frac{3 h^2}{4}}=1 \quad \text{... (ii)}$

Given, $R$ is the point of intersection of tangents at $P$ and at $Q$ .

On solving the equations (i) and (ii)

$\Rightarrow \quad \mathrm{R}=\left(\frac{8}{h}, 0\right)$

Given, $\Delta(h)=$ Area of $\Delta \mathrm{PQR}$

$\begin{aligned}\Rightarrow \quad \Delta(h) & =\frac{1}{2}\left(\frac{8}{h}-h\right) \cdot 2 \sqrt{3-\frac{3 h^2}{4}} \\\Rightarrow \quad \Delta(h) & =\frac{\sqrt{3}}{2}\left(\frac{8}{h}-h\right) \sqrt{4-h^2} \\\Rightarrow \quad \frac{d \Delta(h)}{d h}= & \frac{\sqrt{3}}{2}\left(\frac{-8}{h^2}-1\right) \sqrt{4-h^2} +\frac{\sqrt{3}}{2}\left(\frac{8}{h}-h\right) \cdot \frac{(-2 h)}{2 \sqrt{4-h^2}}\end{aligned}$

$\begin{aligned}\Rightarrow \quad \frac{d \Delta(h)}{d h}= & \frac{-\sqrt{3}}{2 \sqrt{4-h^2}} \\& {\left[\left(\frac{8}{h^2}+1\right)\left(4-h^2\right)+\left(\frac{8}{h}-h\right) h\right] }\end{aligned}$

$\begin{aligned}& \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{-\sqrt{3}}{2 \sqrt{4-h^2}}\left[\frac{32}{h^2}-8+4-\hbar^2+8-h^2\right] \\& \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{-\sqrt{3}}{2 \sqrt{4-h^2}}\left[-2 h^2+\frac{32}{h^2}+4\right] \\& \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{\sqrt{3}}{h^2 \sqrt{4-h^2}}\left[h^4-2 h^2-16\right] \\& \Rightarrow \quad \frac{d \Delta(h)}{d h}=\frac{\sqrt{3}}{h^2 \sqrt{4-h^2}}\left[\left(h^2-1\right)^2-17\right]\end{aligned}$

For, \frac{1}{2} \leq h \leq 1, \quad \frac{d \Delta(h)}{d h}<0

$\therefore \quad \Delta(h)$ is a decreasing function for $x \in\left[\frac{1}{2}, 1\right]$

$\Rightarrow \Delta(1) \leq \Delta(h) \leq \Delta\left(\frac{1}{2}\right)$

Given $\quad \Delta_1=\max _{\frac{1}{2} \leq h \leq 1} \Delta(h)$ and

$\begin{aligned}\Delta_2 & =\min _{\frac{1}{2} \leq h \leq 1} \Delta(h) \\\Rightarrow \quad \Delta_1 & =\Delta\left(\frac{1}{2}\right) \text { and } \Delta_2=\Delta(1) \\\Rightarrow \quad \Delta_1 & =\frac{\sqrt{3}}{2}\left(16-\frac{1}{2}\right) \sqrt{4-\frac{1}{4}} \text { and } \\\Delta_2 & =\frac{\sqrt{3}}{2}(8-1) \sqrt{4-1} \\\Rightarrow \quad \Delta_1 & =\frac{\sqrt{3}}{2} \times \frac{31}{2} \times \frac{\sqrt{5} \cdot \sqrt{3}}{2} \text { and } \Delta_2=\frac{21}{2} \\\Rightarrow \quad \frac{8 \Delta_1}{\sqrt{5}} & =93 \text { and } 8 \Delta_2=84\end{aligned}$

$\Rightarrow \frac{8 \Delta_1}{\sqrt{5}}-8 \Delta_2=9$

Hints:

(i) If $f(x)$ is a decreasing function in $x \in[a, b]$ , then $f(b) \leq f(x) \leq f(a)$

(ii) If $\left(x_1, y_1\right)$ is a point on the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , then equation of tangent at $\left(x_1, y_1\right)$ is $\frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1$ .