Question

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

Answer 1

According to the question, y `dy/dx` = x (where `dy/dx` is the slope of the tangent.)

y dy = x dx

On integrating

⇒ `int y  dy int x  dx`

`y^2/2 = x^2/2 + C`             .... (i)

∵ The curve passes through the point (0, -2)

∴ Putting x = 0, y = - 2,

`4/2 = 0 + C`

⇒  C = 2

On putting C = 2 in equation (i),

`y^2/2 = x^2/2 + 2`   

`=> y^2 = x^2 + 4`

`=> y^2 - x^2 = 4`