Question

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

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

Answer 1

The correct option is D $100$

$\because x^2+4=(sec\theta -cos\theta {)}^2+4=(sec\theta +cos\theta {)}^2....\left(i\right)$
Similarly, $y^2+4=(se{c}^{10}\theta +co{s}^{10}\theta {)}^2....(ii)$
Now, $\frac{dx}{d\theta }=sec\theta tan\theta +sin\theta =tan\theta (sec\theta +cos\theta )$
and $\frac{dy}{d\theta }=10se{c}^9\theta sec\theta tan\theta -10co{s}^9\theta (-sin\theta )$
$=10tan\theta (se{c}^{10}\theta -co{s}^{10}\theta )\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{10tand\theta (se{c}^{10}+co{s}^{10}d\theta )}{tan\theta (sec\theta +cos\theta )}\therefore (\frac{dy}{dx}{)}^2=100\frac{(se{c}^{10}+co{s}^{10\theta })}{(sec\theta +cos\theta {)}^2}=\frac{100(y^2+4)}{(x^2+4)}$
[from Eq. (i), (ii)]
or $(x^2+4)(\frac{dy}{dx}{)}^2=100(y^2+4)$
$k=100$