Question

Find the equation of the curve such that the square of the intercept cut off by any tangent from the y-axis is equal to the product of the co-ordinate of the point of tangent.

Answer 1

Equation of tangent

$(y-y_1)=\frac{dy}{dx}(x-x_1)$

put $x=0\Rightarrow y=\frac{dy}{dx}{x}_1+y_1$

$\Rightarrow y^2=x_1{y}_1\Rightarrow {(\frac{-dy}{dx}{x}_1+y_1)}^2=x_1{y}_1$

$\Rightarrow {\left(x_1\frac{dy}{dx}\right)}^2+y_1^2-2x_1{y}_1\frac{dy}{dx}-x_1{y}_1=0$

$\frac{dy}{dx}=\frac{2x_1{y}_1\pm \sqrt{4x_1^2{y}_1^2-4x_1^2{y}_1^2}}{2x_1^2}=\frac{2x_1{y}_1}{2x_1^2}$

$\frac{dy}{dx}=\frac{y_1}{x_1}$

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

$\Rightarrow lny=lnx+lnc\Rightarrow y=xc$