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The minimum distance between the parabolas $y ^{2} - 4x - 8y + 40 = 0\: and \:x ^{2}- 8x - 4y + 40 = 0$ is
Option: 1
$0$

Option: 2
$\sqrt{3}$

Option: 3
$2\sqrt{2}$

Option: 4
$\sqrt{2}$

Answers (1)

best_answer

Answer (4)

$\\y^{2}-4x-8y+40=0\; \; \; \; ...(i)\\and\; x^{2}-4y-8x+40=0\; \; \; \; \; \; \; ...(ii)$

are mirror image about line $y=x$

So, let equation of tangent parallel to $y=x$ on equation (i) and (ii) is

$\\y=x+\lambda _{1}\; \; \; \; \; \; \; \; \; \; \; ...(iii)\\and \; y=x+\lambda _{2}\; \; \; \; \; \; \; ...(iv)respectively$

Now from (i) and (ii),

$\left ( x+\lambda _{1} \right )^{2}-4\left ( x \right )-8\left (x+\lambda _{1} \right )+40=0$

$\\\Rightarrow x^{2}+x\left ( -12+2\lambda _{1} \right )+\lambda {_{1}}^{2}-8\lambda _{1}+40=0\\D=0$

$\left [ 2\left ( \lambda _{1}-6 \right ) \right ]^{2}-4\left [ \lambda {_{1}}^{2}-8\lambda _{1}+40 \right ]=0$

$\Rightarrow \lambda {_{1}}^{2}+36-12\lambda _{1}-\lambda {_{1}}^{2}+8\lambda _{1}-40=0$

$\Rightarrow -4\lambda _{1}-4=0$

$\lambda _{1}=-1$

Similarly $\lambda _{2}=1$

Now perpendicular distance between

$y=x+1\; and\; y=x-1$

$= \sqrt{2}\; unit$

Posted by

Divya Prakash Singh

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