Question

The points of contact of the tangent drawn from $(0,0)$ 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. None of these

Answer 1

The correct option is C

$x^2–y^2=x^2{y}^2$

The explanation for the correct answer.

Solve for the points of contact of the tangent.

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

The equation is$(Y–y)=\left(\frac{dY}{dx}\right)(X–x)$

$y=\sin x$

Find the first derivative

$$\begin{gathered} \frac{dy}{dx}=\cos x \\ andY–y=\cos x(X–x) \end{gathered}$$

On solving further, as line pass via $(0,0)$

$$\begin{gathered} –y=–x\cos x \\ \Rightarrow \frac{y}{x}=\cos x...\left(i\right) \\ y=\sin x...\left(ii\right) \end{gathered}$$

Square and add the two equations(i) and (ii)

$$\begin{gathered} \left(\frac{y^2}{x^2}\right)+y^2={\cos }^{2}x+{\sin }^{2}x=1 \\ {y}^2+x^2{y}^2=x^2 \\ {x}^2–y^2=x^2{y}^2 \end{gathered}$$

Hence, option(C) is the correct answer.