Question

The points on the parabola $y^2=4x$ which is closest to the circle $x^2+y^2-24y+128=0$ is:

• A. $(0,0)$
• B. $(2,2\sqrt{2})$
• C. $(4,4)$
• D. none of these

Answer 1

Answer: (C)

$(4,4)$

We know that the shortest distance is always along the common normal.

$x^2+y^2-24y+128=0$

Center is $(0,12)$

We have,

$y^2=4x$

Parametric point is $(t^2,2t)$

Differentiate w.r.t $x$

$2y\frac{dy}{dx}=4$

$\frac{dy}{dx}=\frac{2}{y}$

Putting $y=2t$

$\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}$

Slope of the Normal is $=-\frac{dx}{dy}=-t$

Equation of the Normal is :

$y-2t=-t(x-t^2)$

Now this normal must be normal to circle for shortest distance and we know that normal of circle passes through its center.

Put $(0,12)$ in normal

$12-2t=-t(0-t^2)$

$⟹t^3+2t-12=0$

$t=2$

Now we divide the polynomial by $t-2$

$⟹t^3-2t^2+2t^2+2t-12=0$

$⟹t^3-2t^2+2t^2-4t+4t+2t-12=0$

$⟹t^3-2t^2+2t^2-4t+6t-12=0$

$⟹t^2(t-2)+2t(t-2)+6(t-2)=0$

$⟹(t-2)(t^2+2t+6)=0$

As discriminant of $t^2+2t+6=0$ is negative, hence it will have no real roots.

$⟹t-2=0$

$⟹t=2$

Hence the point is,

$(t^2,2t)=(2^2,2\times 2)$

$(4,4)$