Question

Find the equation of a 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

Let x and y be the x-coordinate and y-coordinate of the curve respectively.
We know that the slope of a tangent to the curve in the coordinate axis is given by the $\frac{dy}{dx}$
According to the given information,we get:
$y.\frac{dy}{dx}=x$
$⇒y\, dy=x\, dx$
Integrating both sides,we get:
$∫y\, dy=∫x\, dx$
$⇒\frac{y^2}{2}=\frac{x^2}{2}+C$
$⇒y^2-x^2=2C...(1)$
Now,the curve passes through point(0,-2).
$∴(-2)^2-0^2=2C$
$⇒2C=4$
Substituting $2C=4$ in equation(1),we get:
$y^2-x^2=4$
This is the required equation of the curve.