Question

The minimum distance between the curves $x^2=4y$ and $x^2+y^2+18x+12y+81=0$

• A. $7\sqrt{2}$
• B. $7\sqrt{2}-6$
• C. $7\sqrt{2}+6$
• D. $8$

Answer 1

The correct option is B $7\sqrt{2}-6$

Given, circle is $(x+9{)}^2+(y+6{)}^2=6^2$ and parabola is $x^2=4y$


The minimum distance occurs along the common normal of the two curves.

Equation of the normal will be $x+ty=2t+t^3$
Since, it's passes through $(-9,-6)$
$\therefore -9-6t=2t+t^3\Rightarrow t^3+8t+9=0\Rightarrow (t+1)(t^2-t+8)=0\Rightarrow t=-1$

So, the nearest point of the parabola is $(2t,t^2)\equiv (-2,1)$

So, the distance is $\sqrt{7^2+7^2}-6=7\sqrt{2}-6$