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