Question

The slope of the tangent at any point on a curve is $\lambda$ times the slope of the straight line passing the point of contact to the origin. Find the equation of the curve.

Answer 1

Let $P(x,y)$ be any point on the given curve.

The slope of the tangent at point $P=\lambda$ (Slope of the line joining the point $P$ and the origin)

So,

$\frac{dy}{dx}=\lambda \left(\frac{y-0}{x-0}\right)$

$\Rightarrow \frac{dy}{dx}=\lambda \frac{y}{x}$

$\Rightarrow \frac{dy}{y}=\lambda \frac{dx}{x}$

This is required differential equation,

On integrating and we get,

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

$\log y=\lambda \log x+\log C$

$\Rightarrow \log y=\log x^{\lambda }.C$

$\Rightarrow y=C.x^{\lambda }$

Hence, this is the answer.