The slope of the tangent of any point on a curve is $$\lambda$$ times the slope of the line joining the point of contact to the origin. Formulate the differential equation and hence find the equation of the curve.

Toolbox:
• Slope of the tangent to the curve is $\large\frac{dy}{dx}$
• If the linear differential equation is of the form $\large \frac{dy}{dx}$$=f(x) then it can be solved by seperating the variables and then integrating. • \int\large\frac{dx}{x}=$$\log x$
Step 1:
Let $P(x,y)$ be any point on the given curve.
The slope of the tangent at P=$\lambda$(slope of line joining the point P and the origin)
(i.e) $\large\frac{dy}{dx}=$$\lambda\big(\large\frac{y-0}{x-0}\big) \Rightarrow \large\frac{dy}{dx}=$$\lambda \large\frac{y}{x}$
This is the required differential equation.
$\large\frac{dy}{dx}=\frac{\lambda y}{x}$
Step 2:
Separating the variables we get,
$\large\frac{dy}{y}=$$\lambda \large\frac{dx}{x} Integrating on both sides we get, \int \large\frac{dy}{y}=$$\lambda\int\large\frac{dx}{x}$
(i.e)$\log y=\lambda \log x+\log C$
$\Rightarrow \log y=\log x^{\lambda}.C$
$y=Cx^{\lambda}$
This is the equation of the required curve.