Discussion Forum

1)

A vertical line passing through the point (h,o) 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   $\triangle(h)$  = area of the $\triangle$  PQR , $\triangle_{1}= max \triangle(h)$ .   $\triangle_{2}= min_{1/2\leq h\leq1} \triangle(h)^{1/2\leq h\leq 1}$  , then 

$\frac{8}{\sqrt{5}}\triangle_{1}-8\triangle_{2}$ is equal to 

Answer:

Option A

Explanation:

Concept involved

 As to maximise or minimise area of triangle we sholud find area is terms of parametric coordinate and we second derivative test.

 Here , tangent at   $P(2\cos\theta,\sqrt{3}\sin\theta)$ is  

$\frac{x}{2}\cos\theta+\frac{y}{\sqrt{3}}\sin\theta=1$

 $\therefore$    $R(\sec\theta,0)$

$\Rightarrow$     $\triangle= area of \triangle PQR$

  $=\frac{1}{2}(2\sqrt{3}\sin\theta)(2\sec\theta-2\cos\theta)$

    $=2\sqrt{3}.\sin^{3}\theta\cos\theta$   .........(i)

  Since,   $\frac{1}{2}\leq h\leq1,$

 $\therefore$      $\frac{1}{2}\leq2 cos \theta\leq1,$

    $\Rightarrow$    $\frac{1}{4}\leq cos \theta\leq\frac{1}{2}$

$\therefore$   $\frac{d\triangle}{d\theta}=\frac{2\sqrt{3}\left\{\cos\theta.3\sin^{3}\theta \cos\theta-\sin^{3}\theta(-\sin \theta)\right\}}{\cos^{2}\theta}$

    =   $\frac{2\sqrt{3}\sin^{2}\theta}{\cos^{2}\theta}[3\cos^{2}\theta+ \sin^{2}\theta]$

    =   $\frac{2\sqrt{3}\sin^{2}\theta}{\cos^{2}\theta}[2\cos^{2}\theta+ 1]$

    =  $2\sqrt{3}\tan^{2}\theta(2\cos^{2}\theta+1)>0$

   when, $\frac{1}{4}\leq \cos \theta\leq\frac{1}{2}$

$\therefore$   $\triangle_{1}=\triangle_{max}$   occurs at cos $\theta$ = $\frac{1}{4}$

=  $\left(\frac{2\sqrt{3}\sin^{3}\theta}{\cos\theta}\right)$

when   $\cos\theta=\frac{1}{4}=\frac{45\sqrt{5}}{8}$

$\triangle_{2}=\triangle_{min}$   occurs at cos $\theta$ = $\frac{1}{2}$

 =  $\left(\frac{2\sqrt{3}\sin^{3}\theta}{\cos\theta}\right)$

 when    $\cos\theta=\frac{1}{2}=\frac{9}{2}$

$\therefore$   $\frac{8}{\sqrt{5}}\triangle_{1}-8\triangle_{2}=45-36=9$