Question

Tangents are drawn from the origin to the curve y = sin x, then their point of contact lie on the curve

• A. $x^2+y^2=1$
• B. $x^2-y^2=1$
• C. $\frac{1}{x^2}+\frac{1}{y^2}=1$
• D. $\frac{1}{y^2}-\frac{1}{x^2}=1$

Answer 1

The correct option is D

$\frac{1}{y^2}-\frac{1}{x^2}=1$

The equation of tangent at (x,y) is
Y - y = cos x (X - x)
It pass through (0, 0)
Then, o - y = cos x ( 0 - x)
Or $cosx=\frac{y}{x}$
Given, sin x = y
Squaring and adding Eqs. (i) and (ii), then
$1=\frac{y^2}{x^2}+y^2\Rightarrow \frac{1}{y^2}-\frac{1}{x^2}=1$