Question

The points of contact of the tangents drawn from the origin to the curve $y=\sin x$ lie on the curve

• A. $x^2-y^2=xy$
• B. $x^2+y^2=x^2{y}^2$
• C. $x^2-y^2=x^2{y}^2$
• D. $Noneofthese$

Answer 1

Answer: (C)

$x^2-y^2=x^2{y}^2$

Let the tangent be drawn at the point $(x,y).$

Its equation is $Y-y=\frac{dy}{dx}(X-x)$

But $y=\sin x$

$\therefore \frac{dy}{dx}=\cos x$

$\Rightarrow Y-y=\cos x(X-x)$

Since it passes through $(0,0)$, therefore substituting $(x,y)$ by $(0,0)$ we get

$-y=-x\cos x\Rightarrow \frac{y}{x}=\cos x$ and $y=\sin x$

$\therefore \frac{y^2}{x^2}+y^2={\cos }^{2}x+{\sin }^{2}x=1$

$\Rightarrow y^2+x^2{y}^2=x^2\Rightarrow x^2-y^2=x^2{y}^2$

hence the points of contact lie on $x^2-y^2=x^2{y}^2$