Review question

# Where does this tangent to this ellipse meet $x=a$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9325

## Solution

The point $P$ in the first quadrant lies on the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. The points $A^{\prime} \ (-a, 0)$ and $A \ (a, 0)$ are the extremities of the major axis of the ellipse, and $O$ is the origin. The tangent to the ellipse at $P$ meets the axis of $y$ at $Q$ and meets the line $x = a$ at $T$. The chord $A^{\prime} P$ meets the axis of $y$ at $M$, and when produced meets the line $x=a$ at $R$.

Prove that

1. $AT=TR$,

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 $AT = TR$.

1. $OQ^2 - MQ^2 = b^2$.

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 $OQ^2 - MQ^2 = b^2$.