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AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Section: Mathematics

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Question : 66 of 160

Marks: +1, -0

If x = sec θ - cos θ , y =

\sec^{10} \theta - \cos^{10} \theta

(x^2+4)\left(\frac{dy}{dx}\right)^2

(y^2+4)

, then k =

$\frac{1}{100}$
1
10
100

Solution:

It is given that,

x=\sec \theta - \cos \theta

y=\sec^{10} \theta - \cos^{10} \theta

And

(x^{2}+4)\left(\frac{dy}{dx}\right)^{2}=k(y^{2}+4)

Therefore

\frac{dy}{d\theta}=10(\sec^{10}\theta+\cos^{10}\theta)\tan\theta

And

\frac{dx}{d\theta}=(\sec\theta+\cos\theta)\tan\theta

Hence,

\frac{dy}{dx}=10\frac{\sec^{10}\theta+\cos^{10}\theta}{\sec\theta+\cos\theta}

\left(\frac{dy}{dx}\right)^{2}=100\left(\frac{\sec^{10}\theta+\cos^{10}\theta}{\sec\theta+\cos\theta}\right)^{2}

=100\left(\frac{\sec^{20}\theta+\cos^{20}\theta+2}{\sec^{2}\theta+\cos^{2}\theta+2}\right)

=100\left(\frac{(\sec^{10}\theta-\cos^{10}\theta)^{2}+4}{(\sec\theta-\cos\theta)^{2}+4}\right)

Solve further

\left(\frac{dy}{dx}\right)^{2}=100\left(\frac{y^{2}+4}{x^{2}+4}\right)

(x^{2}+4)\left(\frac{dy}{dx}\right)^{2}=100(y^{2}+4)

=k(y^{2}+4)

Therefore

k=100

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