A curve passes through the point $(0,-2)$ and at any point $(x, y)$ of 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. Find the equation of the curve.

Home » A curve passes through the point and at any point of the curve, the product of the slope of its tangent and -coordinate of the point is equal to the -coordinate of the point. Find the equation of the curve.

A curve passes through the point $(0,-2)$ and at any point $(x, y)$ of 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. Find the equation of the curve.

A curve passes through the point $(0,-2)$ and at any point $(x, y)$ of 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. Find the equation of the curve.

Solution:

It is given that product of slope of tangent and $y$ coordinate equals the $x$ -coordinate i.e., $y \frac{d y}{d x}=x$
We have, $y d y=x d x$
$\begin{array}{l} \Rightarrow \int y d y=\int x d x \\ \Rightarrow \frac{y^{2}}{2}=\frac{x^{2}}{2}+c \end{array}$
For the curve passes through $(0,-2)$ , we get $c=2$
Hence, the required particular solution is:-
$\therefore y^{2}=x^{2}+4$