Shortest distance between the parabola y² = 4x and x² + y² – 4x – 16y + 64 = 0 is equal to

The correct option/answer is 2√5 – 2.

To find the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2–4x–16y+64=0$, we must analyze the geometry of both curves.

First, let us simplify the equation of the circle to find its center and radius. The given circle equation is:
$x^2+y^2–4x–16y+64=0$
We can rewrite this by completing the square for both $x$ and $y$ terms:
$(x^2–4x+4)+(y^2–16y+64)–4–64+64=0$
${(x–2)}^2+{(y–8)}^2=4$
Comparing this with the standard equation of a circle ${(x–h)}^2+{(y–k)}^2=r^2$, we find:
Center of the circle, $C=(2,8)$
Radius of the circle, $r=2$

The shortest distance between a curve and a circle lies along the common normal. Specifically, the shortest distance from the circle to the parabola is equal to the shortest distance from the center of the circle to the parabola, minus the radius of the circle.

Let any parametric point on the parabola $y^2=4x$ (where $a=1$) be represented as $P(t^2,2t)$.

The equation of the normal to the parabola $y^2=4x$ at the point $(t^2,2t)$ is:
$y=–tx+2t+t^3$

For this normal to pass through the center of the circle $C(2,8)$, we substitute $x=2$ and $y=8$ into the normal equation:
$8=–t\left(2\right)+2t+t^3$
$8=–2t+2t+t^3$
$t^3=8$
Taking the real cube root, we get:
$t=2$

Substituting $t=2$ back into the coordinates of point $P$:
$P=(2^2,2(2\left)\right)=(4,4)$

Now, we calculate the distance between the center of the circle $C(2,8)$ and the point $P(4,4)$ using the distance formula:
$CP=\sqrt{{(4–2)}^2+{(4–8)}^2}$
$CP=\sqrt{2^2+{(–4)}^2}$
$CP=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$

The shortest distance $d$ between the parabola and the circle is the distance $CP$ minus the radius $r$ of the circle:
$d=CP–r$
$d=2\sqrt{5}–2$

Therefore, the shortest distance is indeed $2\sqrt{5}–2$.