Question

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

• A. $(2,4)$
• B. $(2,-4)$
• C. $(18,-12)$
• D. $(8,8)$

Answer 1

Answer: (B)

$(2,-4)$

let $(a,b)$ is the point on the parabola from which the distance between given curve is minimum
$\Rightarrow b^2=8a$ and centre of circle is $(0,-6)$
assume d is the distance
$d^2=a^2+(b+6{)}^2=b^4/64+(b+6{)}^2$
for minimum value of $d$
$\frac{d\left(d^2\right)}{d{b}^2}=0=b^3/16+2(b+6)$
$\Rightarrow b^3+32b+192=0\Rightarrow (b+4)(b^2-4b+48)=0$
$\Rightarrow b=-4$ is only root
and $a=b^2/8=16/8=2$
Hence require point on the parabola is $(2,-4)$