Question

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

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

Answer 1

The correct option is C $(2,-4)$

A point on the parabola is at minimum distance from the circle if and only if it is at minimum distance from the centre of the circle.
Any point on the parabola $y^2=8x$ is of the form $P(2t^2,4t)$
The centre of the circle $x^2+{(y+6)}^2=1$ is $O(0,-6)$
${OP}^2=4t^4+{(-6-4t)}^2=4(t^4+4t^2+12t+9)$
Let $A=t^4+4t^2+12t+9$
$\Rightarrow \frac{dA}{dt}=4t^3+8t+12$ $=4(t^3+2t+3)=4(t+1)(t^2-t+3)$

so, $\frac{dA}{dt}=0$ if $t=-1$,

Moreover at $t=-1$; $\frac{d^{2}A}{d{t}^2}=4(3{(-1)}^2+2)>0$ which represents minima.
Hence, required point is $P(2,-4)$

Hence, option C.