Question

Tangents are drawn from origin to the curve $y=\sin x$ then point of contect lies on -

• A. $x^2=y^2$
• B. $x^2{y}^2=0$
• C. $x^2{y}^2=x^2-y^2$
• D. None of these

Answer 1

Answer: (C)

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

$y=\sin x$
$\Rightarrow \frac{dy}{dx}=\cos x$
Let the point of contact of tangent passing through origin to the given curve is $P(h,k)$
So equation of tangent to this curve passes through $(0,0)$ is,
$y=\left(\cos h\right)x$
For point of contact this line will pass through point P,
$\Rightarrow k=h\cos h\Rightarrow \cos h=\frac{k}{h}...\left(1\right)$
Also point P lies on the given curve,
$\sin h=k..\left(1\right)$
Now using $(1{)}^2+(2{)}^2=1$
$\frac{k^2}{h^2}+k^2=1\Rightarrow k^2{h}^2=h^2-k^2$
Thus locus of $P(h,k)$ is,
$x^2{y}^2=x^2-y^2$
Hence, option 'C' is correct.