Question

The parametric equations of a curve are given by $x={\sec }^{2}t,y=\cot t.$ Tangent at $Pt=\frac{\pi }{4}$ meets the curve again at Q; then $PQ=?$

• A. $\left(2\sqrt{5}\right).$
• B. $\frac{\left(5\sqrt{5}\right)}{2}.$
• C. $\frac{\left(3\sqrt{5}\right)}{2}.$
• D. $\left(3\sqrt{5}\right).$

Answer 1

The correct option is C $\frac{\left(3\sqrt{5}\right)}{2}.$

$x={\sec }^{2}t,y=\cot t.$
Given $t=\frac{\pi }{4}$
$\Rightarrow x={\sec }^2\left(\frac{\pi }{4}\right),y=\cot \frac{\pi }{4}$
$\Rightarrow x=2,y=1$
So, the point P is $(2,1)$
Now, $\frac{dx}{dt}=2\sec t\sec t\tan t$
$\frac{dy}{dt}=-cose{c}^{2}t$
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}{\cot }^{3}t$
Slope of tangent at P is $-\frac{1}{2}$
Equation of tangent at P is
$y-1=-\frac{1}{2}(x-2)$
$\Rightarrow x+2y=4$ .....(1)
Since, ${\sec }^{2}t-{\tan }^{2}t=1$
$\Rightarrow x-\frac{1}{y^2}=1$ .....(2)
Solving (1) and (2), we get
$(4-2y)y^2-1=y^2.$
$\Rightarrow y=1,1,-\frac{1}{2}$
$\Rightarrow x=2,5$ (by (1))
$\therefore Qis(5,-\frac{1}{2})$
$\therefore PQ=\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}.$
The values of $y=1,1$ correspond to point P.