Find the equation of a curve passing through the point (0, –2) given that at any point 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.
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.