Question

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

A
5
B
21
C
285
D
215

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 $$=2\sqrt { 5 } -\sqrt { 5 } =\sqrt { 5 }$$.Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More