Question

A curve is passing through the point $(1,2)$ and if the perpendicular from $(0,0)$ to the tangent at any point on the curve is equal to the abscissa of the point of contact, then the sum of length of $x$-intercept and $y$-intercept of the curve is

Answer 1

$\text{Let the point of contact be}P(x,y)\text{Equation of tangent}:y_1-y=m(x_1-x)\Rightarrow m{x}_1-y_1-mx+y=0\text{Perpendicular distance from}(0,0)=x\Rightarrow \left|\frac{0-0-mx+y}{\sqrt{m^2+1}}\right|=x\Rightarrow y^2-x^2=2mxy\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}[\because m=\frac{dy}{dx}]\text{Put}y=vx\text{and proceed},v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\int \frac{2v}{v^2+1}dv=-\int \frac{dx}{x}\Rightarrow \ln (v^2+1)=-\ln x+\ln c\Rightarrow v^2+1=\frac{c}{x}\Rightarrow y^2+x^2=cx[\because y=vx]\text{Now, it passes through}(1,2)\Rightarrow c=5\Rightarrow y^2+x^2-5x=0\text{For}x\text{-intercept, put}y=0\Rightarrow x=0,5\Rightarrow \text{Length of}x\text{-intercept}=5\text{For}y\text{-intercept, put}x=0\Rightarrow y=0\Rightarrow \text{Length of}y\text{-intercept}=0\therefore \text{Sum of length of both intercept}=5$