Animated Solution for Mathematics - Conic Sections: Let $y=mx+c,m>0$ be the focal chord of $y^2=-64x$, which is tangent to $(x+10{)}^2+y^2=4$. Then, the value of $4\sqrt{2}(m+c)$ is equal to

Imagine you are standing on a coordinate plane, looking at two distinct geometric entities: a parabola and a circle. The parabola, defined by $y^2=-64x$, is a wide, sweeping curve opening to the left.

The circle, defined by $(x+10{)}^2+y^2=4$, is a small, compact shape centered near the origin. The problem asks us to find a line that acts as a bridge between these two—a focal chord of the parabola that is also a tangent to the circle.

First, we must understand our parabola. By comparing $y^2=-64x$ with the standard form $y^2=4ax$, we immediately see that $4a=-64$, which means $a=-16$.

The focus of this parabola is at $(a,0)$, or $(-16,0)$. This point is the heart of the parabola. Any focal chord, by definition, must pass through this point.

This is our first constraint: our line $y=mx+c$ must contain the point $(-16,0)$.

Since the line passes through $(-16,0)$, we substitute these coordinates into the line equation $y=mx+c$. This gives us $0=m(-16)+c$, leading to the elegant relationship $c=16m$.

Now, our line equation $y=mx+c$ transforms into $y=mx+16m$, or more conveniently:

We have successfully reduced our line to a single variable, $m$, the slope.

Now, let us look at the circle $(x+10{)}^2+y^2=4$. This is a circle centered at $(-10,0)$ with a radius $r=\sqrt{4}=2$.

The problem states that our focal chord is tangent to this circle. In the language of geometry, this means the perpendicular distance from the center of the circle to the line must be exactly equal to the radius of the circle.

We use the perpendicular distance formula:

Here, the center $(x_1,y_1)$ is $(-10,0)$, and the line is $mx-y+16m=0$. Substituting these values, we get:

Simplifying the numerator, we get $|6m|$, and the denominator is $\sqrt{m^2+1}$. So:

Dividing both sides by 2, we arrive at $|3m|=\sqrt{m^2+1}$.

Squaring both sides gives $9m^2=m^2+1$, which simplifies to $8m^2=1$, or $m^2=\frac{1}{8}$. Since the problem specifies $m>0$, we have $m=\frac{1}{2\sqrt{2}}$.

With $m$ in hand, we find $c=16m=16\times \frac{1}{2\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}$.

Finally, we calculate the target expression $4\sqrt{2}(m+c)$. Substituting our values:

The beauty of this problem lies in how the variables cancel out, leaving us with a clean, integer result of 34.