**Question:**

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.

**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.

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