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Question

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


A
0
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B
3
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C
22
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D
2
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Solution

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 $$y=x$$

$$\implies y=(1)x$$

$$\therefore Slope=1$$

Minimum distance (points are common tangent and common normal)

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

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

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

$$y=6$$

Putting $$y$$ in parabola $$01$$

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

$$\implies 4x=28$$

$$\implies x=7$$

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

Maths

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