Question

The minimum distance between the parabolas $y^2-4x-8y+40=0$ and $x^2-8x-4y+40=0$ is

Answer 1

The equations of the parabolas are $y^2-4x-8y+40=0\cdots \left(1\right)$
$x^2-8x-4y+40=0\cdots \left(2\right)$
We observe that if we interchange $x$ and $y$ in equation $\left(1\right)$, we obtain equation $\left(2\right)$
So, the two parabolas are symmetric about $y=x$.
Shortest distance exist between the tangents on both the parabolas which are parallel to $y=x$

For curve $1$, the equation of tangent is
$y{y}_1-2(x+x_1)-4(y+y_1)+40=0$
$\Rightarrow -2x+(y_1-4)y-2x_1-4y_1+40=0$
Slope $=\frac{2}{y_1-4}=1$
$\Rightarrow y_1=6$
Putting value of $y_1$ in eqn $\left(1\right)$, we get $x_1=7$
So, $(x_1,y_1)=(7,6)$
Since, the parabolas are symmetric about $y=x$, and if $(x_1,y_1)=(7,6)$, then $(x_2,y_2)=(6,7)$
$(\because x$ and $y$ will interchange$)$

$\therefore d=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=\sqrt{2}$