Question

Find the equation of the curve passing through the point (0,-2) given that at any point (x,y) on the curve the product of the slope of its tangent and y coordinate of the point is equal to the x-coordinate of the point.

Answer 1

Slope of the tangent at any point (x,y) on the curve is given by $\frac{dy}{dx}$.

$y\frac{dy}{dx}=x$

$ydy=xdx$

Integrating both sides, we get,

$\int ydy=\int xdx$

$\frac{y^2}{2}=\frac{x^2}{2}+C$

When, $x=0,y=-2$

$\frac{4}{2}=C$

$\Rightarrow C=2$

Hence, equation of the required curve is

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

$y^2=x^2+4$