The shortest distance between the curves $y^2=8 \mathrm{x}$ and $x^2+y^2+12 y+35=0$ is :

Question

Aicharya · Accepted Answer

Equation of normal to parabola
$y^2=8 x \text { is } y=m x-4 m-2 m^3$
passes through $(0,-6)$ we get
$\begin{aligned}
& -6=-4 \mathrm{~m}-2 \mathrm{~m}^3 \\ & \Rightarrow \mathrm{~m}^3+2 \mathrm{~m}-3=0 \\ & \Rightarrow(\mathrm{~m}-1)\left(\mathrm{m}^2+\mathrm{m}+3\right)=0 \Rightarrow \mathrm{~m}=-1 \\ & \mathrm{P}=\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(2,-4) \\ & \therefore \text { Shortest distance }=\mathrm{PC}-\mathrm{r} \\ & =(2 \sqrt{2}-1)
\end{aligned}$

The shortest distance between the curves $y^2=8 \mathrm{x}$ and $x^2+y^2+12 y+35=0$ is :

Solution