Question

A curve passes through (3,-4). Slope of tangent at any point (x, y) is 2y/x.Find the equation of the curve.

Answer 1

Answer:

$\bf\:4x^2+9y=0$

Step-by-step explanation:

A curve passes through (3,-4). Slope of tangent at any point (x, y) is 2y/x.Find the equation of the curve.

$\text{slope of tangent at any point}=\frac{dy}{dx}$

$\frac{2y}{x}=\frac{dy}{dx}$

separating the variables, we get

$\frac{dy}{y}=2\frac{dx}{x}$

Integrating on both sides, we get

$\int\frac{dy}{y}=2\int\frac{dx}{x}$

$log\:y=2\:log\:x+logc$

$log\:y=log\:x^2+logc$

$log\:y=log\:x^2c$

$y=x^2c$

It passes through (3,-4)

$-4=3^2c$

$c=\frac{-4}{9}$

The equation of the required curve is

$y=x^2(\frac{-4}{9})$

$9y=-4x^2$

$\implies\:\boxed{\bf\:4x^2+9y=0}$