1.

Find the equation of the normal to the curve x=12 cosec⁡θ and y=2 sec⁡θ at x=π/4 .(a) \(\frac{1}{6}\)(b) -6(c) 6(d) –\(\frac{1}{6}\)I got this question during an internship interview.My question comes from Derivatives Application in section Application of Derivatives of Mathematics – Class 12

Answer»

The correct choice is (c) 6

To explain: GIVEN that, x=12 cosecθ and y=2 sec⁡θ

\(\frac{dx}{dθ}\)=-12 COSEC θ cot⁡θ

\(\frac{dy}{dθ}\)=2 tan⁡θ sec⁡θ

\(\frac{dy}{dx}\)=\(\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{2 \,tan⁡θ \,sec⁡θ}{-12 \,cosec θ \,cot⁡θ}\)=-\(\frac{1}{6} \frac{sin⁡θ}{cos^2⁡θ} × \frac{cos⁡θ}{sin^2⁡⁡θ}\) = –\(\frac{cot⁡θ}{6}\)

\(\frac{dy}{dx}]_{x=\frac{\pi}{4}}\)=\(-\frac{\frac{cot\pi}{4}}{6}=-\frac{1}{6}\)

Hence, the slope of NORMAL at θ=π/4 is

–\(\frac{1}{slope \,of \,tangent \,at \,θ=\pi/4}=-\frac{-1}{\frac{1}{6}}\)=6


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