Question

$A$ curve is given by the equations $x={\sec }^2\theta ,y=\cot \theta$. If the tangent at $P$ where $\theta =\frac{\pi }{4}$ meets the curve again at $Q$, then length $PQ$, is

• A. $\frac{\sqrt{5}}{2}$
• B. $\frac{3\sqrt{5}}{2}$
• C. $10$
• D. $20$

Answer 1

The correct option is B $\frac{3\sqrt{5}}{2}$

$y=\cot \theta$

$\Rightarrow \frac{dy}{d\theta }=-{\text{cosec}}^2\theta$

$x={\sec }^2\theta$
$\Rightarrow \frac{dx}{d\theta }=2\sec \theta \cdot \sec \theta \tan \theta$

${\frac{dy}{dx}|}_{\theta =\frac{\pi }{4}}=-\frac{{\text{cosec}}^2\theta }{2{\sec }^2\theta \tan \theta }$
$=\frac{-2}{2\times 2\times 1}=-\frac{1}{2}$

$y=\frac{1}{\tan \theta }$
$x=1+{\tan }^2\theta$
$x=1+\frac{1}{y^2}$
$\Rightarrow y^{2}x=y^2+1$
$\Rightarrow x=2,y=1$
Now, at $(x,y)=(2,1)$, equation of a line with slope $-\frac{1}{2}$, is

$(y-1)=-\frac{1}{2}(x-2)$
$\Rightarrow 2y+x=4$
$\Rightarrow y^2(4-2y)=y^2+1$
$\Rightarrow 4y^2-2y^3=y^2+1$
$\Rightarrow 2y^3-3y^2+1=0$
$\Rightarrow y=1,-\frac{1}{2}$
$\Rightarrow x=2,5$

$PQ=\sqrt{(2-5{)}^2+{(1+\frac{1}{2})}^2}$
$={(3^2+\frac{3^2}{4})}^{0.5}$
$=\frac{3\sqrt{5}}{2}$

Hence, option B.