Animated Solution for Mathematics - Conic Sections: Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then

Welcome, fellow explorers of the mathematical universe! Today, we are embarking on a journey through a classic coordinate geometry problem that bridges the gap between the elegance of parabolas and the symmetry of circles.

Imagine you are standing at the center of a circle, looking out at a parabola. You want to reach that parabola as quickly as possible. What is the shortest path? This is the core of our problem today.

Decoding the Curves

First, let us look at our players. We have a parabola defined by $y^2=4x$. This is a standard right-opening parabola with its vertex at the origin and $a=1$.

Then, we have a circle defined by the general equation $x^2+y^2-4x-16y+64=0$. To make sense of this circle, we need to bring it into its standard form.

By completing the square for the $x$ and $y$ terms, we get:

This simplifies beautifully to:

Now, the geometry becomes clear: we have a circle centered at $S(2,8)$ with a radius $r=2$.

The Normal's Secret

Here is the crucial insight: the shortest distance from a point to a curve is always measured along the normal to the curve at that point. If you were to draw a line from the center $S$ to the parabola, the shortest segment would be the one that hits the parabola at a right angle.

This line is the normal to the parabola. To find this, we use the parametric form of the normal for the parabola $y^2=4ax$.

With $a=1$, any point $P$ on the parabola can be represented as $(t^2,2t)$. The equation of the normal at this point is given by:

This equation is our key to unlocking the location of $P$.

Solving for the Parameter

We know that this normal line must pass through our center $S(2,8)$. So, we substitute $x=2$ and $y=8$ into our normal equation:

Look at that! The $2t$ terms on both sides cancel out perfectly, leaving us with $t^3=8$. Taking the cube root, we find $t=2$.

This is the magic value that defines our point $P$. Substituting $t=2$ back into our parametric coordinates $P(t^2,2t)$, we find:

Verifying the Options

Now that we have $P(4,4)$ and $S(2,8)$, we can calculate the distance $SP$ using the distance formula:

This confirms that Option A is correct!

Next, let us look at the normal's $x$-intercept. With $t=2$, the normal equation becomes $y+2x=2(2)+2^3$, or $y+2x=12$.

To find the $x$-intercept, we set $y=0$, which gives $2x=12$, so $x=6$. Option C is also correct!

Finally, let us consider the tangent at $Q$. Point $Q$ lies on the circle and on the segment $SP$. The slope of the normal $SP$ is:

Since the tangent at $Q$ is perpendicular to the radius $SQ$ (which lies on the normal $SP$), the slope of the tangent must be the negative reciprocal of $-2$, which is $\frac{1}{2}$. Thus, Option D is correct.

This problem is a beautiful reminder that when we break down complex curves into their fundamental properties—normals, tangents, and centers—the most daunting equations often collapse into simple, elegant solutions. Keep practicing, and keep visualizing!