Let a line perpendicular to the line 2 x-y=10 touch the parabola y^2=4(x-9) at the point P…

Question

Let a line perpendicular to the line $2 x-y=10$ touch the parabola $y^2=4(x-9)$ at the point P. The distance of the point P from the centre of the circle $x^2+y^2-14 x-8 y+56=0$ is \_\_\_\_\_\_\_\_\_\_.

Answer: 10

Accepted Answer

Line perpendicular to $2x-y=10$ have slope $=\frac{-1}{2}$

$\Rightarrow$ Line tangent to parabola ${y}^{2}=4(x-9)$ with slope $m$ is

$\begin{gathered} y=m(x-9)+\frac{1}{m},m=\frac{-1}{2} \ \Rightarrow y=\frac{-(x-9)}{2}-2\Rightarrow 2y=-x+9-4 \ \Rightarrow 2y+x=5 \end{gathered}$

Solving the tangent and parabola we get point $P$

$\begin{gathered} {(\frac{5-x}{2})}^{2}=4(x-9)\Rightarrow {x}^{2}-10x+25=16x-144 \ \Rightarrow {x}^{2}-26x+169=0\Rightarrow (x-13{)}^{2}=0 \ \Rightarrow P\equiv (13,-4) \end{gathered}$

Distance of $P$ from the centre of circle $(7,4)$ is $\sqrt{(13-7{)}^{2}+(-4-4{)}^{2}}=\sqrt{36+64}=10$ units.

Solutions · 1