Question

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

• A. $0$
• B. $\sqrt{3}$
• C. $2\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

The parabolas $y^2-4x-8y+40=0$ and $x^2-8x-4y+40=0$ are symmetric about $y=x$

$⟹y=\left(1\right)x$

$\therefore Slope=1$

Minimum distance (points are common tangent and common normal)

$2y\frac{dy}{dx}-4-8\frac{dy}{dx}+0=0$

$\frac{dy}{dx}(y-4)-2=0$

$(y-4)=2\left[\frac{dy}{dx}\right(slope=1\left)\right]$

$y=6$

Putting $y$ in parabola $01$

$36-4x-48+40=0$

$⟹4x=28$

$⟹x=7$

Points on parabola $01=(7,6)$

Points on parabola $02=(6,7)$

Since it is symmetric to $y=x$

Distance between them is

$=\sqrt{(7-6{)}^2+(6-7{)}^2}$

$=\sqrt{1+(-1{)}^2}$

$=\sqrt{2}$ units