Question

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

• A. $(2,-4)$
• B. $(18,-2)$
• C. $(2,4)$
• D. none of these

Answer 1

Answer: (A)

$(2,-4)$

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 $\frac{dD}{dt}=0$

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

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

Hence the required point is $\left(2\right(1{)}^2,4\left(1\right))=(2,-4)$