Parabolas Inscribed in Shared Tangent Lines

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Theorem

Let the function $\map f x = A x^2 + B x + C_1$ be a curve embedded in the Euclidean Plane.

Let $\map {y_1} x$ be the equation of the tangent line at $\tuple {Q, \map f Q}$ on $f$.

Let $\map {y_2} x$ be the equation of the tangent line at $\tuple {-Q, \map f {-Q} }$ on $f$.


Then there exists another function $\map g x$ also embedded in the Euclidean Plane defined as:

$\map g x = -A x^2 + B x + C_2$.

with:

tangent lines $\map {y_3} x$ being the equation of the tangent line at $\tuple {Q, \map g Q}$ on $g$

and:

$\map {y_4} x$ being the equation of the tangent line at $\tuple {-Q, \map g {-Q} }$ on $g$.

so that the tangent lines $y_3$ and $y_4$ inscribe $\map f x$ and the tangent lines $y_1$ and $y_2$ inscribe $\map g x$.


Proof

The tangent line at $\tuple {Q, \map f \Q}$ on $f$ is defined as:

$\map {y_1} x = \paren {2 A Q + B} x + b_1$

where $2 A Q + B$ is the slope of the tangent line on the point $\tuple {Q, \map g Q}$ on $f$.

Substitute in the coordinates of the point $\tuple {Q, \map g Q}$ to $y_1$ and solve for $b_1$.

This will reveal the $y$-intercept of $y_1$:

\(\ds A Q^2 + B Q + C_1\) \(=\) \(\ds \paren {2 A Q + B} Q + b_1\) The value of $y_1 = \map f Q$
\(\ds A Q^2 + B Q + C_1 -2 A Q^2 - B Q\) \(=\) \(\ds b_1\)
\(\ds -A Q^2 + C_1\) \(=\) \(\ds b_1\)


Continue by following the same steps for $y_2$ which is defined:

$\map {y_2} x = \paren {-2 A Q + B} x + b_2$

where $-2 A Q + B$ is the slope of the tangent line at the point $\tuple {-Q, \map f {-Q} }$ on $f$.

Substitute in the coordinates of the point $\paren {-Q, \map f {-Q} }$ to $y_2$.

Use these values to solve for $b_2$, and this will reveal the $y$-intercept of $y_2$:

\(\ds A \paren {-Q}^2 + B \paren {-Q} + C_1\) \(=\) \(\ds \paren {-2 A Q + B} \paren {-Q} + b_2\) the value of $y_2 = \map f {-Q}$
\(\ds A Q^2 - B Q + C_1 -2 A Q^2 + B Q\) \(=\) \(\ds b_2\)
\(\ds -A Q^2 + C_1\) \(=\) \(\ds b_2\)

The $y$-intercepts of both $y_1$ and $y_2$ have been shown to be equivalent.

$\Box$


Since $b_1 = b_2$ redefine this value as $b$.

The distance between $b$ and $C_1$ is $\size {C_1 - b}$.

Let $\map g x = -A x^2 + B x + C_2$.

Then the tangent line at the point $\tuple {Q, \map g Q}$ on $g$ is defined as:

$\map {y_3} x = \paren {-2 A Q + B} x + b_3$

where $-2 A Q + B$ is the slope of the tangent line at $\tuple {Q, \map g Q}$ on $g$.

Solve for $b_3$ using the same methods used for $y_1$ and $y_2$.

This will reveal the $y$-intercept of $y_3$:

$b_3 = A Q^2 + C_2$

The result also follows for the tangent line $\tuple {-Q, \map g {-Q} }$ on $g$ which is defined:

$y_4 = \paren {-2 A Q + B} x + b_4$

Solving for $b_4$ yields the result:

$b_4 = A Q^2 + C_2$

The $y$-intercepts of both $y_3$ and $y_4$ have been shown to be equivalent.

$\Box$


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

\(\ds \map {g'} Q\) \(=\) \(\ds \map {f'} {-Q}\)
\(\ds \map {g'} {-Q}\) \(=\) \(\ds \map {f'} Q\)

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.

$\Box$


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

Solve for $C_2$ to determine the vertical translation of $\map g x$:

\(\ds C_2\) \(=\) \(\ds \paren {-A Q^2 + C_1} - A Q^2\)
\(\ds C_2\) \(=\) \(\ds -2 A Q^2 + C_1\)

Therefore the function:

$\map g x = -A x^2 + B x - \paren {2 A Q^2 + C_1}$

will have tangent lines equivalent to the tangent lines on $\map f x$ at the points $\tuple {Q, \map f Q}$, and $\tuple {-Q, \map f {-Q} }$.

$\blacksquare$

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