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

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

Question

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. $AT=TR$,

---

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

Suggestion

How can we find the tangent to the ellipse? Be careful when you differentiate the equation of the ellipse!

Do not forget that we know that the point $P$ lies on the ellipse.

Solution

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

$$\begin{equation} \frac{p^2}{a^2} + \frac{q^2}{b^2} = 1. \end{equation}$$

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

$(ellipse-eqn-eval-at-P)$

, the tangent is described by

$$\begin{equation} y = - \frac{p b^2}{q a^2} x + \frac{b^2}{q}. \end{equation}$$

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

$$\begin{equation} y = \frac{q}{p + a} x + \frac{q a}{p + a}. \end{equation}$$

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

$(ellipse-eqn-eval-at-P)$

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

((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

((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

$(ellipse-eqn-eval-at-P)$

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