Find the equation of a curve passing through the point

Solution:

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 relation,

$\frac{d y}{d x}$

According to the given information, we get:

$y \cdot \frac{d y}{d x}=x$

$\Rightarrow y d y=x d x$

Integrating both sides, we get:

$\int y d y=\int x d x$

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

$\Rightarrow y^{2}-x^{2}=2 \mathrm{C}$                  ...(1)

Now, the curve passes through point (0, –2).

$\therefore(-2)^{2}-0^{2}=2 C$

$\Rightarrow 2 C=4$

Substituting 2C = 4 in equation (1), we get:

$y^{2}-x^{2}=4$

This is the required equation of the curve.