Let a line perpendicular to the line $2x-y=10$ touch the parabola $y^2=4(x-9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2+y^2-14x-8y+56=0$ is
Step-by-Step Solution
Key Concept: Line perpendicular to $2x-y=10$ has slope $-1/2$. Parabola $y^2=4(x-9)$: $a=1$, vertex $(9,0)$. Tangent with slope $m$: $y=m(x-9)+1/m$. For slope $m=-1/2$: tangent $y=-\frac{1}{2}(x-9)-2$.
$P=(13,-4)$, centre $(7,4)$. Distance $=10$.
Correct Answer: 10