Parabolas Inscribed in Shared Tangent Lines

Theorem
Let $f \left({x}\right) = A x^2 + B x + C_1$.

Let $y_1 \left({x}\right)$ be the equation of the tangent line at $\left({Q, f \left({Q}\right)}\right)$ on $f$.

Let $y_2 \left({x}\right)$ be the equation of the tangent line at $\left({-Q, f \left({-Q}\right)}\right)$ on $f$.

Then there exists a function:
 * $g \left({x}\right) = -A x^2 + B x + C_2$.

With $y_{3}\left(x\right)$ being the equation of the tangent line at $\left(Q, g\left(Q\right)\right)$ on $g$.

With $y_{4}\left(x\right)$ being the equation of the tangent line at $\left(-Q, g\left(-Q\right)\right)$ on $g$.

So that the tangent lines $y_{3}$ and $y_{4}$ inscribe $f\left(x\right)$ and the tangent lines $y_{1}$ and $y_{2}$ inscribe $g\left(x\right)$.

Proof
The tangent line at $\left(Q, f\left(Q\right)\right)$ on $f$ is defined:


 * $y_{1}\left(x\right) = \left(2AQ+B \right)x+b_{1}$

where $2AQ+B$ is the slope of the tangent line on the point $\left(Q, f \left(Q \right)\right)$ on $f$. Sub in the coordinates of the point $\left(Q, f \left(Q \right)\right)$ to $y_{1}$ and solve for $b_{1}$.

This will reveal the $y$-intercept of $y_{1}$:

Continue by following the same steps for $y_{2}$ which is defined:


 * $y_{2}\left(x\right) = \left(-2AQ+B\right)x+b_{2}$

where $-2AQ+B$ is the slope of the Tangent line at the point $\left(-Q, f\left(-Q\right)\right)$ on $f$.

Sub in the coordinates of the point $\left(-Q, f\left(-Q\right)\right)$ to $y_{2}$.

Use these values to solve for $b_{2}$;

this will reveal the $y$-intercept of $y_{2}$:

The $y$-intercepts of both $y_{1}$ and $y_{2}$ have been shown to be equivalent.

Since $b_{1} = b_{2}$ redefine this value as $b$.

The distance between $b$ and $C_{1}$ is $|C_{1}-b|$.

Let $g\left(x\right)=-Ax^2+Bx+C_{2}$;

then the Tangent line at the point $\left(Q, g\left(Q\right)\right)$ on $g$ is defined:


 * $y_{3}(x)=(-2AQ+B)x+b_{3}$

where $-2AQ+B$ is the slope of the Tangent line at $\left(Q, g\left(Q\right)\right)$ on $g$.

Solve for $b_{3}$ using the same methods used for $y_{1}$ and $y_{2}$.

This will reveal the $y$-intercept of $y_{3}$:


 * $b_{3} = AQ^2+C_{2}$

The result also follows for the Tangent line $\left(-Q, g\left(-Q\right)\right)$ on $g$ which is defined:


 * $y_{4} = \left(-2AQ+B\right)x+b_{4}$

Solving for $b_{4}$ yields the result:


 * $b_{4} = AQ^2+C_{2}$

The $y$-intercepts of both $y_{3}$ and $y_{4}$ have been shown to be equivalent.

Notice that the derivatives of $g$ and $f$ satisfy:

Then it must be true that $y_{1}=y_{4}$ and $y_{2}=y_{3}$,

and the functions $y_{1}$, $y_{2}$, $y_{3}$, and $y_{4}$ share the same $y$-intercept.

Redefine this the $y$-intercepts of the tangent lines as $b$.

Solve for $C_{2}$ to determine the vertical translation of $g\left(x\right)$:

Therefore the function:


 * $g(x) = -Ax^2+Bx-\left(2AQ^2+C_{1}\right)$

will have Tangent lines equivalent to the Tangent lines on $f\left(x\right)$ at the points $\left(Q, f\left(Q\right)\right)$, and $\left(-Q, f\left(-Q\right)\right)$.