Question

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