If $x=\sec heta-\cos heta, y=\sec ^{10} heta-\cos ^{10} heta$ and $\left(x^2+4 ight)\left(\frac{d y}{d x} ight)^2=k\left(y^2+4 ight)$, then $k=$

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If $x=\sec 	heta-\cos 	heta, y=\sec ^{10} 	heta-\cos ^{10} 	heta$ and $\left(x^2+4ight)\left(\frac{d y}{d x}ight)^2=k\left(y^2+4ight)$, then $k=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Given, } y=\sec ^{10} 	heta-\cos ^{10} 	heta 	ext { and } x=\sec 	heta-\cos 	heta \ & 	ext { so, } \quad \frac{d y}{d 	heta}=10\left(\sec ^9 	heta \sec 	heta 	an 	heta+\cos ^9 	heta \sin 	hetaight) \ & =10\left(\sec ^{10} 	heta+\cos ^{10} 	hetaight) 	an 	heta \ & 	ext { and } \frac{d x}{d 	heta}=\sec 	heta 	an 	heta+\sin 	heta=(\sec 	heta+\cos 	heta) 	an 	heta \ & 	herefore \quad \frac{d y}{d x}=10 \frac{\sec ^{10} 	heta+\cos ^{10} 	heta}{\sec 	heta+\cos 	heta} \ & \Rightarrow \quad\left(\frac{d y}{d x}ight)^2=100\left(\frac{\sec ^{10} 	heta+\cos ^{10} 	heta}{\sec 	heta+\cos 	heta}ight)^2 \ & \end{aligned}$ $ \begin{aligned} & =100 \frac{\sec ^{20} 	heta+\cos ^{20} 	heta+2}{\sec ^2 	heta+\cos ^2 	heta+2} \ & =100 \frac{\left(\sec ^{10} 	heta-\cos ^{10} 	hetaight)^2+4}{(\sec 	heta-\cos 	heta)^2+4} \ & =100\left(\frac{y^2+4}{x^2+4}ight) \ \Rightarrow \quad\left(x^2+4ight)\left(\frac{d y}{d x}ight)^2 & =100\left(y^2+4ight) \ & =K\left(y^2+4ight) \end{aligned} $ (given) So, $K=100$

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

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