Question

The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact. Let the equation of the curve satisfying the above condition and which passes through (1 , 1) be $k{x}^2+m{y}^2-2nx=0$.Find $k+m+n$?

Answer 1

Let the point of contact be P(x,y).
Equation of tangent at P(x,y) is given by
$Y-y=\frac{dy}{dx}(X-x)$
Now given , length of perpendicular from origin to tangent $=x$
$\Rightarrow \frac{|-x\frac{dy}{dx}+y|}{\sqrt{1+(\frac{dy}{dx}{)}^2}}=x$
$\Rightarrow x^2(\frac{dy}{dx}{)}^2+y^2-2xy\frac{dy}{dx}=x^2[1+(\frac{dy}{dx}{)}^2]$
$\Rightarrow y^2-2xy\frac{dy}{dx}=x^2$
$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$
$\Rightarrow x\frac{dv}{dx}=-\frac{1+v^2}{2v}$
$\Rightarrow \frac{2v}{1+v^2}=-\frac{dx}{x}$
Integrating both sides, we get
$\log (1+v^2)+\log x=\log k$
$\log x(1+\frac{y^2}{x^2})=\log k$
$\Rightarrow x^2+y^2=kx$
Since, it passes through (1,1)
$\Rightarrow k=2$
So, the equation of curve is
$x^2+y^2-2x=0$
On comparing with the given equation,
$k=1,m=1,n=-1$
$\Rightarrow k+m-n=3$