Question

If $x=sec\theta -cos\theta ,y=se{c}^{10}\theta -co{s}^{10}\theta$ and $(x^2+4){\left(\frac{dy}{dx}\right)}^2=k(y^2+4)$, then k is equal to

• A. $\frac{1}{100}$
• B. $1$
• C. $10$
• D. $100$

Answer 1

The correct option is D $100$

$\because x^2+4=(sec\theta -cos\theta {)}^2+4=(sec\theta +cos\theta {)}^2\cdots \left(i\right)Similarly,y^2+4=(se{c}^{10}\theta +co{s}^{10}\theta {)}^2\cdots (ii)Now,\frac{dx}{d\theta }=sec\theta tan\theta +sin\theta =tan\theta (sec\theta +cos\theta )and\frac{dy}{d\theta }=10se{c}^9\theta sec\theta tan\theta -10co{s}^9\theta (-sin\theta )=10tan\theta (se{c}^{10}\theta -co{s}^{10}\theta )\Rightarrow \frac{dy}{dx}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=\frac{10tan\theta (se{c}^{10}\theta +co{s}^{10}\theta )}{tan\theta (sec\theta +cos\theta )}\therefore {\left(\frac{dy}{dx}\right)}^2=100\frac{(se{c}^{10}\theta +co{s}^{10}\theta )}{(sec\theta +cos\theta {)}^2}=\frac{100(y^2+4)}{(x^2+4)}or(x^2+4){\left(\frac{dy}{dx}\right)}^2=100(y^2+4)$
On comparing with the expression given we get k = 100