Question

Find the equation of the curve passing through the (0, –2) given that at any point (x, y) on the curve the product of the slope of its tangent and y-co-ordinate of the point is equal to the x-co-ordinate of the point.

Options

• `y = x^2`

• `y^2 - x^2` = 4

• `y^2 = x^2 + 4`

• Both `y^2 - x^2 = 4` and `y^2 = x^2 + 4`

Answer 1

Both `y^2 - x^2 = 4` and `y^2 = x^2 + 4`

Explanation:

Slope tangent to the curve at `(x, y) = (dy)/(dx)`

We are given `((dy)/(dx)) = x`

∴ `ydx = xdx`

Integrating we get, `y^2/2 = x^2/2 + c`

or `y^2 = x^2 + 2c`

The curve passes through (0, –2)

∴ 4 = 0 + 2c

∴ 2c = 4

∴ Equation of the curve is `y^2 = x^2 + 4` or `y^2 = x^2 = 4`