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  4. 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=π / 4 meets the curve again at the point Q, then |PQ| is equal to

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Q. 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 $t=\pi / 4$ meets the curve again at the point $Q$, then $|PQ|$ is equal to

Application of Derivatives

Solution:

image
Eliminating $t$ gives $y^{2}(x-1)=1$
Equation of the tangent at $P(2,1)$ is $x+2 y=4$,
Solving with curve $x=5$ and $y=-1 / 2,$ we get
$Q \equiv(5,-1 / 2)$ or $PQ=\frac{3 \sqrt{5}}{2}$