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 which of the following is/are correct

• A. $PQ=\frac{3\sqrt{5}}{2}$
• B. $PQ=\frac{5\sqrt{3}}{2}$
• C. Slope of tangent at $Q=\frac{1}{16}$
• D. Slope of tangent at $Q=-\frac{1}{16}$

Answer 1

Answer: (A), (C)

Slope of tangent at $Q=\frac{1}{16}$

Eliminating $t$ we will get curve equation as $y^2(x-1)=1.$
Differentiating we will get
$2y{y}^{\prime }(x-1)+y^2=0$
$\frac{dy}{dx}=\frac{-y}{2(x-1)}$
At $(2,1),m=-\frac{1}{2}$
Equation of tangent at $P(2,1)$ is
$y-1=-\frac{1}{2}(x-2)$
$x+2y=4.$
$\Rightarrow x=4-2y$
Substituting $x$ value in curve equation, we get
$y^2(3-2y)=1$
$2y^3-3y^2+1=0$
$\Rightarrow y=1,\frac{-1}{2}$
$Q=(5,-\frac{1}{2})$
$\Rightarrow PQ=\frac{3\sqrt{5}}{2}$ units and
Slope of tangent at $Q=\frac{1}{2\times 2\times 4}=\frac{1}{16}$