Question

A curve is represented by the equations, $x={\sec }^{2}t$ and $y=\cot t$, where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t=\frac{\pi }{4}$ meets the curve again at the point $Q$ then $|PQ|$ is equal to

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

Answer 1

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

$x={\sec }^{2}t$
$y=\cot t$
$\frac{dy}{dx}=-\frac{{\csc }^{2}t}{2{\sec }^{2}t\cdot \tan t}=-\frac{{\cot }^{3}t}{2}$

Now,
${\frac{dy}{dx}|}_{t={45}^0}=-\frac{1}{2}$

${x|}_{t={45}^0}=2$
${y|}_{t={45}^0}=1$

Hence the equation of the tangent will be
$y-1=-\frac{1}{2}(x-2)\Rightarrow 2y-2=-x+2\Rightarrow x+2y=4$ ...(i)

Now,
${\sec }^{2}t=1+{\tan }^{2}t$
${\sec }^{2}t=1+\frac{1}{{\cot }^{2}t}$
Or
$x=1+\frac{1}{y^2}\Rightarrow x{y}^2=y^2+1$ is the equation of curve.

Solving the equation of tangent and equation of curve we get
$4-2y=1+\frac{1}{y^2}$
Or
$4y^2-2y^3=y^2+1\Rightarrow 2y^3-3y^2+1=0$
$y=-\frac{1}{2}$ and $y=1$

For $y=1;x=2$

For $y=-\frac{1}{2};x=5$
Hence,

$Q=(5,-\frac{1}{2})$ and $P=(2,1)$
$\therefore PQ=\frac{3\sqrt{5}}{2}$.