Question

A tangent at a point P on the curve cuts the x - axis at A and B is the foot of perpendicular from P on the x axis. If the midpoint of AB is fixed at $(\alpha ,0)$ for any point P, find the differential equation and hence find the curve .

Answer 1

Let $y=f\left(x\right)$

$Y-y=\frac{dy}{dx}(X-x)$

$-y=\frac{dy}{dx}(X-x)$ at $y=0$

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

C is the midpoint of AB $(\alpha ,0)=(\frac{x+x-y\frac{dx}{dy}}{2},0)$

$\Rightarrow 2\alpha =2x-y\frac{dx}{dy}$

$\Rightarrow y\frac{dx}{dy}=2(x-\alpha )$ is the required D.E.

$\Rightarrow \int \frac{dx}{2(x-\alpha )}=\int \frac{dy}{y}$

$\Rightarrow \frac{1}{2}ln(x-\alpha )=lny+c$

$\Rightarrow ln(x-\alpha {)}^{1/2}-lny=c$

$\Rightarrow ln\left(\frac{\sqrt{x-\alpha }}{y}\right)=c$

$\Rightarrow \sqrt{x-\alpha }=y{e}^c$

$\Rightarrow x-\alpha =y^2{e}^{2c}$

$\Rightarrow x=\alpha +y^2{e}^{2c}$.