Where does this tangent to this ellipse meet $x=a$?

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Let us denote the coordinates of \(P\) by \((p, q)\).

Since \(P\) is in the first quadrant we have \(p, q > 0\), and since \(P\) lies on the ellipse we know that \[\begin{equation} \frac{p^2}{a^2} + \frac{q^2}{b^2} = 1. \label{eq:ellipse-eqn-eval-at-P} \end{equation}\] Let us first work out the equation for the tangent to the ellipse at \(P\) in the form \(y = mx + c\). Differentiating implicitly we have \[\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0\] and hence \(\dfrac{dy}{dx} = - \dfrac{x b^2}{y a^2}\). The value of \(m\) is the value of this derivative at the point \((p, q)\), which is \(m = - \dfrac{p b^2}{q a^2}\). We also know that the point \(P\) lies on the tangent and hence we have the condition \[q = - \frac{p^2 b^2}{q a^2} + c\] which is equivalent to \[c = \frac{q^2 a^2 + p^2 b^2}{q a^2}.\] Using \(q^2 a^2 + p^2 b^2 = a^2 b^2\), which follows directly from the above equation \(\eqref{eq:ellipse-eqn-eval-at-P}\), the tangent is described by \[\begin{equation} y = - \frac{p b^2}{q a^2} x + \frac{b^2}{q}. \label{eq:equation-of-tangent-at-P} \end{equation}\]

Let us now work out the coordinates of the point \(T\) shown in the following picture.

We can get the \(y\)-coordinate by inserting \(x = a\) into the equation of the tangent at \(P\). We get \[y = - \frac{p b^2}{q a^2}a + \frac{b^2}{q} = \frac{a b^2 - p b^2}{q a}\] and hence \(T\) has the coordinates \(\left(a, \dfrac{a b^2 - p b^2}{q a}\right)\).

We will now calculate the coordinates of the point \(R\) drawn in the following picture.

We start off by computing the equation for the line \(A^{\prime} P\). Since the coordinates of \(P\) are \((p, q)\), the slope of the line is given by \(\dfrac{q}{p + a}\) and hence the line is of the form \(y = \dfrac{q}{p + a} x + c\) for the \(y\)-axis intercept \(c\). Using that the point \(A^{\prime}\) lies on the line we get the condition \(- \dfrac{q a}{p + a} + c = 0\) and hence \(c = \dfrac{q a}{p + a}\). Thus the line \(A^{\prime} P\) has the equation \[\begin{equation} y = \frac{q}{p + a} x + \frac{q a}{p + a}. \label{eq:equation-Aprime-P} \end{equation}\]

Since the \(x\)-coordinate for \(R\) is \(a\) we can compute the \(y\)-coordinate via \[y = \frac{q a}{p + a} + \frac{q a}{p + a}.\] and hence the point \(R\) has the coordinates \(\left(a, 2\dfrac{q a}{p + a}\right)\).

Since the points \(A\), \(T\), and \(R\) all lie on the same vertical line \(x = a\) it suffices to show that \(T\) lies exactly in the middle of \(A\) and \(R\), i.e. that the \(y\)-coordinate of \(T\) is half the \(y\)-coordinate of \(R\). From equation \(\eqref{eq:ellipse-eqn-eval-at-P}\) we get \(b^2 = \dfrac{q^2 a^2}{a^2 - p^2}\). Thus we have \[\begin{align*} b^2 \frac{a - p}{q a} &= \dfrac{q^2 a^2}{a^2 - p^2} \frac{a - p}{q a} \\ &= \frac{qa(a-p)}{a^2 - p^2} \\ &= \frac{qa(a-p)}{(a+p)(a-p)} \\ &= \frac{qa}{a+p} \end{align*}\]

and hence \(AT = TR\).

Let us now compute the coordinates of the points \(Q\) and \(M\) shown in the following drawing.

The \(y\)-coordinate of \(Q\) is given by the \(y\)-axis intercept of the tangent to the ellipse at \(P\). We have already found the intercept as part of equation \(\eqref{eq:equation-of-tangent-at-P}\), so we already know that the point \(Q\) has the coordinates \(\left(0, \dfrac{b^2}{q}\right)\). Similarly, the \(y\)-coordinate of \(M\) is just the \(y\)-axis intercept of the line \(A^{\prime} P\), i.e. the line given by equation \(\eqref{eq:equation-Aprime-P}\), and hence \(M\) is given by \(\left(0, \dfrac{q a}{p + a}\right)\). Using this we get \[\begin{align*} OQ^2 - MQ^2 &= \left( \frac{b^2}{q} \right) ^2 - \left( \frac{b^2}{q} - \frac{q a}{p + a} \right)^2 \\ &= \left( \frac{b^2(p + a)}{q (p + a)} \right)^2 - \left( \frac{b^2(p + a) - q^2 a}{q (p + a)} \right)^2 \\ &= \frac{2q^2 a b^2 (p + a) - q^4 a^2}{q^2 (p + a)^2} \\ &= \frac{2a b^2 (p + a) - q^2 a^2}{(p + a)^2} \end{align*}\] Equation \(\eqref{eq:ellipse-eqn-eval-at-P}\) implies that \(q^2 = b^2 \left(1 - \dfrac{p^2}{a^2}\right)\) and using this for our calculation gives \[\begin{align*} OQ^2 - MQ^2 &= \frac{2a b^2 (p + a) - b^2 \left(1 - \frac{p^2}{a^2}\right) a^2}{(p + a)^2} \\ &= b^2 \frac{2a (p + a) - a^2 + p^2}{(p + a)^2} \\ &= b^2 \frac{2ap + 2a^2 - a^2 + p^2}{(p + a)^2} \\ &= b^2 \underbrace{\frac{p^2 + 2p a + a^2}{(p + a)^2}}_{= 1}. \end{align*}\]

Thus \(OQ^2 - MQ^2 = b^2\).