Animated Solution for Mathematics - Conic Sections: A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^2=2x$ and touches the parabola $y=(x-\frac{1}{4}{)}^2+\alpha$, where $\alpha >0$. Then $(4\alpha -8{)}^2$ is equal to ____.

We begin by dissecting the first parabola, $y^2=2x$. This is a classic, right-opening parabola. By comparing it to the standard form $y^2=4ax$, we identify $4a=2$, which gives us $a=1/2$.

This tells us the vertex is at $V(0,0)$ and the focus is at $S(1/2,0)$. We introduce a circle with a radius $R=2$ that passes through both these points.

Because the circle passes through $(0,0)$ and $(1/2,0)$, the segment connecting them is a chord of the circle. The center of the circle must lie on the perpendicular bisector of this chord.

Since the chord lies on the x-axis, its perpendicular bisector is the vertical line $x=1/4$. This is our first major breakthrough: the x-coordinate of the circle's center, $h$, must be $1/4$.

With the x-coordinate $h=1/4$ in hand, we can find the y-coordinate, $k$. We know the circle passes through the origin $(0,0)$. The equation of the circle is $(x-h{)}^2+(y-k{)}^2=R^2$.

Substituting our known values, we get:

This simplifies to $1/16+k^2=4$. Solving for $k^2$, we find:

Thus, $k=\pm \frac{\sqrt{63}}{4}$. For our visualization, we work with the positive value, $k=\frac{\sqrt{63}}{4}$. We have now completely defined the circle's position, centered at $(1/4,\frac{\sqrt{63}}{4})$ with a radius of $2$.

A Dance of Symmetry

Now, consider the second parabola, $y=(x-1/4{)}^2+\alpha$. This parabola is also symmetric about the line $x=1/4$.

The vertex of this second parabola is at $(1/4,\alpha )$. Since both the circle and the parabola share the same axis of symmetry, $x=1/4$, they must touch at the vertex of the parabola.

For the circle to touch the parabola from the outside, the vertex of the parabola must be at the top of the circle. The distance from the center of the circle $(1/4,k)$ to the vertex of the parabola $(1/4,\alpha )$ must be exactly the radius of the circle, $R=2$.

Therefore, $\alpha -k=2$, or $\alpha =k+2$.

We have $\alpha =k+2$. The question asks us to find the value of $(4\alpha -8{)}^2$. Let us substitute our expression for $\alpha$:

Rearranging this, we get $4\alpha -8=4k$. We already know $k=\frac{\sqrt{63}}{4}$.

So, $4k=4(\frac{\sqrt{63}}{4})=\sqrt{63}$. Now, squaring both sides, we get:

The final answer is 63. The complexity of the problem melts away when you see the symmetry. We started with two seemingly disconnected parabolas and a circle, and through the power of coordinate geometry, we found that they were perfectly aligned.