Question

# The coordinates of the point on the parabola $$y^{2}=8x$$ which is at minimum distance from the circle $$x^{2}+\left ( y+6 \right )^{2}=1$$ are

A
(2,4)
B
(18,12)
C
(2,4)
D
none of these

Solution

## The correct option is A $$\left ( 2,-4 \right )$$Let the parametric point on the parabola be $$(2t^{2},4t)$$.Now the centre of the circle lies at $$(0,-6)$$ and has a radius equal to 1.Hence the distance of the centre of the circle from parametric point is $$OP^{2}$$ $$=(0-2t^{2})^{2}+(-6-4t)^{2}$$$$=4t^{4}+16t^{2}+48t+36$$$$=4[t^{4}+4t^{2}+12t+9]$$Hence, $$\dfrac{d(OP^{2})}{dt}$$ $$=4[4t^{3}+8t+12]$$ $$=0$$$$t^{3}+2t+3=0$$$$t=-1$$ is the only real root.And $$\dfrac{d^{2}(OP^{2})}{dt^{2}}_{t=-1}>0$$Now we get the parametric point as $$(2,-4)$$Hence the required point is $$(2,-4)$$.Mathematics

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