Question

Tangents are drawn from origin to the curve $y=\sin x+\cos x.$ Then their points of contact lie on the curve.

• A. $\frac{1}{x^2}+\frac{2}{y^2}=1$
• B. $\frac{2}{x^2}-\frac{1}{y^2}=1$
• C. $\frac{2}{x^2}+\frac{1}{y^2}=1$
• D. $\frac{2}{y^2}-\frac{1}{x^2}=1$

Answer 1

The correct option is D $\frac{2}{y^2}-\frac{1}{x^2}=1$

​​​​​​$y=\sin x+\cos x$
$\Rightarrow y=\sqrt{2}\sin (x+\frac{\pi }{4})$
$\frac{dy}{dx}=\sqrt{2}\cos (x+\frac{\pi }{4})$
Given line passes through origin so it will be in form $y=mx\Rightarrow m=\frac{y}{x}$
$\Rightarrow \frac{dy}{dx}=\frac{y_1}{x_1}=\sqrt{2}\cos (x_1+\frac{\pi }{4})$ where $(x_1,y_1)$ is point on the curve
$\therefore \frac{y_1^2}{x_1^2}=2{\cos }^2(x_1+\frac{\pi }{4})=2(\frac{-y_1^2}{2}+1)$
$\Rightarrow$Locus of $(x_1,y_1)$ is $\frac{2}{y^2}-\frac{1}{x^2}=1$