Question

The locus of the foot of the perpendicular drawn from origin to a straight line which passes through a fixed point $P(h,k)$ is denoted by S. The tangent at $P(h,k)$ on the curve S is given by

• A. $xh+yk-k^2=0$
• B. $xh+yk-h^2=0$
• C. $2xh+2yk-h^2-k^2=0$
• D. $xh+yk-h^2-k^2=0$

Answer 1

The correct option is D $xh+yk-h^2-k^2=0$

The equation of a variable line passing through $P(h,k)$ is given by $y-k=m(x-h)$
If $(\alpha ,\beta )$ is the foot of the perpendicular drawn from the origin to the line, then
$\frac{\beta -0}{\alpha -0}\times m=-1$
$m=\frac{-\alpha }{\beta }$
$\Rightarrow \beta -k=m(\alpha -h)$
$\beta -k=\frac{-\alpha }{\beta }(\alpha -h)$
$\Rightarrow {\alpha }^2+{\beta }^2-\alpha h-\beta k=0$

$x^2+y^2-xh-yk=0$ is locus which is a circle.
By point form, tangent equation is
$xh+yk-h\left(\frac{x+h}{2}\right)-k\left(\frac{y+k}{2}\right)=0$
$xh+yk-h^2-k^2=0$ is the equation of tangent