Question

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

• A. $x^2{y}^2=y^2-x^2$
• B. $x^2{y}^2=y^2+x^2$
• C. $x^2{y}^2=x^2-y^2$
• D. None of these

Answer 1

Answer: (C)

Let $\left(x_1,y_1\right)$ be one of the points of contact.
Given curve is $y=\cos x$
$\Rightarrow \frac{dy}{dx}=-\sin x$
$\Rightarrow {\left|\frac{dy}{dx}\right|}_{\left(x_1,y_1\right)}=-\sin x_1$
Now the equation of the tangent at $\left(x_1,y_1\right)$ is
$y-y_1{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\left(x-x_1\right)$ $\Rightarrow y-y_1=-\sin x_1\left(0-x_1\right)$
Since, it is given that equation of tangent passes through origin
$\Rightarrow 0-y_1=-\sin x_1\left(0-x_1\right)$
$\therefore y_1=-x_1\sin x_1$ ... (1)
also, point $\left(x_1,y_1\right)$ lies on
$\therefore y_1=\cos x_1$
From Eqs (i), (ii) we get
${\sin }^2{x}_1+{\cos }^2{x}_1=\frac{y_1^2}{x_1^2}+y_1^2=1$
$\Rightarrow x_1^2=y_1^2+y_1^2{x}_1^2$
Hence, the locus of $x^2=y^2+y^2{x}^2$
$\Rightarrow x^2{y}^2=x^2-y^2$