Question

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

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

Answer 1

Answer: (D)

Given, $y={\sec }^{10}\theta -{\cos }^{10}\theta$ and $x=\sec \theta -\cos \theta$
So, $\frac{dy}{d\theta }=10\left({\sec }^9\theta \sec \theta \tan \theta +{\cos }^9\theta \sin \theta \right)$
$=10\left({\sec }^{10}\theta +{\cos }^{10}\theta \right)\tan \theta$
and $\frac{dx}{d\theta }=\sec \theta \tan \theta +\sin \theta =(\sec \theta +\cos \theta )\tan \theta$
$\therefore \frac{dy}{dx}=10\frac{{\sec }^{10}\theta +{\cos }^{10}\theta }{\sec \theta +\cos \theta }$
$\Rightarrow {\left(\frac{dy}{dx}\right)}^2=100{\left(\frac{{\sec }^{10}\theta +{\cos }^{10}\theta }{\sec \theta +\cos \theta }\right)}^2$
$=100\frac{{\sec }^{20}\theta +{\cos }^{20}\theta +2}{{\sec }^2\theta +{\cos }^2\theta +2}$
$=100\frac{{\left({\sec }^{10}\theta -{\cos }^{10}\theta \right)}^2+4}{(\sec \theta -\cos \theta {)}^2+4}$
$=100\left(\frac{y^2+4}{x^2+4}\right)$
$\Rightarrow \left(x^2+4\right){\left(\frac{dy}{dx}\right)}^2=100\left(y^2+4\right)$
$=K\left(y^2+4\right)$ (given)
$So,K=100$