Single Choice

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

A$$(2,-4)$$
Correct Answer
B$$(18,-2)$$
C$$(2,4)$$
Dnone of these

Solution

Centre of the given circle is, $$C(0,-6)$$

Let any point on the given parabola $$y^2=8x$$ is $$(2t^2,4t)$$

Now let the distance of this point from the centre of the circle is $$d$$

$$\therefore d^2=(2t^2)^2+(4t+6)^2=D$$ (say)

Thus for minimum distance $$\cfrac{dD}{dt}=0$$

$$\Rightarrow 16t^3+8(4t+6)=0\Rightarrow t^3+2t+3=0\Rightarrow (t+1)(t^2-t+3)=0$$

$$\Rightarrow t=-1$$ is only real solution

Hence the required point is $$(2(1)^2,4(1))=(2,-4)$$

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