Question

Tangents are drawn to the curve $y=\sin x$ from the origin. The point of contact lie on

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

Answer 1

The correct option is B $x^2{y}^2=x^2-y^2$

Points on the curve $y=\sin x$ that have tangent lines through the origin satisfy:

$slopeofcurveat(x,y)$ is $\frac{y}{x}=\frac{\sin x}{x}.$

By simple differentiation, the slope of the curve at $x$ is $\cos x$, so our points satisfy

$\cos x=\frac{\sin x}{x}$

$\therefore x=\tan x$ or $ta{n}^{-1}x=x$

So we have

$y=\sin x=\sin \left(ta{n}^{-1}x\right)=\frac{x}{\sqrt{x^2+1}}$

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

$y^2(x^2+1)=x^2\Rightarrow x^2{y}^2=x^2-y^2$

Hence, $\left(B\right)$