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Question

Minimum distance between the curves $${ y }^{ 2 }=4x$$ and $${ x }^{ 2 }+{ y }^{ 2 }-12x+31=0$$ is


A
5
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B
21
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C
285
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D
215
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Solution

The correct option is A $$\sqrt { 5 } $$
Centre and radius of the given circle is $$P\left( 6,0 \right) $$ and $$\sqrt { 5 }$$, respectively.

Now the minimum distance between two curves always occurs along a line that normal to both the curves.

Equation of normal to $${ y }^{ 2 }=4x$$ at $$\left( { t }^{ 2 },2t \right) $$ is

$$y=-tx+2t+{ t }^{ 3 }$$.

If it is normal to circle also, then it must pass through $$\left( 6,0 \right)$$.

$$\therefore 0={ t }^{ 3 }-4t\Rightarrow t=0$$   or $$t=\pm 2$$.

$$\Rightarrow A\left( 4,4 \right) $$ and $$C\left( 4,-4 \right) $$

$$\Rightarrow PA=PC=\sqrt { 20 } =2\sqrt { 5 } $$

Required minimum distance $$=2\sqrt { 5 } -\sqrt { 5 } =\sqrt { 5 } $$.

729601_670287_ans_cb56cdc060984ac788f0dff8dc1acb52.png

Mathematics

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