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.

Answer 1

Let ( x,y ) be the point of abscissa and ordinate respectively. Slope of the tangent is given in the form of differential as dy dx = m tangent

According to the question it is given that,

y⋅ dy dx =x ydy=xdx

Integrate both sides,

∫ ydy = ∫ xdx y 2 2 = x 2 2 +C y 2 2 − x 2 2 =C

y 2 − x 2 =2C(1)

Equation (1) is the required equation of curve that passes through point ( 0,−2 ).

( −2 ) 2 −0=2C 2C=4

Substitute the value in equation (1).

y 2 − x 2 =4

Thus, required equation of the curve that passes through ( 0,−2 ) is y 2 − x 2 =4.