Animated Solution for Mathematics - Conic Sections: The equations of the common tangents to the parabola $y=x^2$ and $y=-(x-2{)}^2$ is/are

Imagine you are standing on a coordinate plane. In front of you, two parabolas are dancing.

The first, $P_1:y=x^2$, is a classic, smiling upward from the origin $(0,0)$.

The second, $P_2:y=-(x-2{)}^2$, is a frown, opening downwards with its peak at $(2,0)$. They are locked in a geometric embrace, and our mission is to find the lines that touch both of them—the common tangents.

Our First Tool

To find a line that touches both, we start by defining a general line: $y=mx+c$. Here, $m$ is the slope and $c$ is the y-intercept.

For this line to be tangent to $P_1$, it must intersect the parabola at exactly one point. Substituting $y=mx+c$ into $y=x^2$ yields:

For this to be a tangent, the discriminant $D$ must be zero. Calculating $D=(-m{)}^2-4(1)(-c)=m^2+4c$, we set it to zero to find:

This is our golden key: any line of the form $y=mx-\frac{m^2}{4}$ is guaranteed to be tangent to our first parabola.

Bridging the Gap

Now, we apply this tool to the second parabola, $P_2:y=-(x-2{)}^2$. We require this same line to be tangent to $P_2$ as well.

Substituting our tangent equation into the second parabola's equation:

Expanding the right side gives $mx-\frac{m^2}{4}=-(x^2-4x+4)$, which simplifies to $mx-\frac{m^2}{4}=-x^2+4x-4$. Rearranging everything to one side, we obtain the quadratic:

For this line to be tangent to the second parabola, the discriminant of this new quadratic must also be zero. We calculate:

Expanding this, we get $(m^2-8m+16)-(16-m^2)=0$. The $16$s cancel out, leaving us with:

Factoring this, we find $2m(m-4)=0$. This gives us two solutions: $m=0$ and $m=4$.

When $m=0$, our intercept $c$ becomes $0$, giving us the tangent $y=0$ (the x-axis).

When $m=4$, our intercept $c$ becomes $-\frac{4^2}{4}=-4$, giving us the tangent $y=4x-4$, or $y=4(x-1)$.