Review question

# What can we say if a point $P$ on an ellipse is directly above the focus? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6710

## Solution

The point $S$ is a focus of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, and $P$ is a point on the ellipse such that $PS$ is perpendicular to the axis of $x$. The tangent and normal to the ellipse at $P$ meet the axis of $y$ in $Q$ and $R$ respectively. If $H$ is the other focus of the ellipse prove that $QR=HP$.

Firstly, we can work out the coordinates of $P$.

It is a standard property of ellipses that the distance from the centre $O$ to either focus is $f=\sqrt{a^2-b^2}$ (with the convention that $a>b$). Thus at points $S$ and $P$ we have $x=\sqrt{a^2-b^2}$. Thus the $y$-coordinate of $P$ is \begin{align*} \frac{y^2}{b^2} &= 1-\frac{a^2-b^2}{a^2} \\ &= \frac{b^2}{a^2}, \end{align*}

so $y^2 = \frac{b^4}{a^2}$ and so $y = \frac{b^2}{a},$ since we have picked $P$ to be in the top right quadrant. Thus $P$ has coordinates $\left(\sqrt{a^2-b^2},\dfrac{b^2}{a}\right)$.

Now let’s work out the equation of the tangent to the ellipse at $P$.

We differentiate the equation of the ellipse implicitly to find the gradient of the tangent at $P$: we have $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0,$ and so $\frac{dy}{dx} = -\frac{2x}{a^2}\times \frac{b^2}{2y} = -\frac{x}{y}\times\frac{b^2}{a^2}.$ Thus at point $P$ the derivative is \begin{align*} \frac{dy}{dx} &= -\frac{\sqrt{a^2-b^2}}{b^2/a}\times\frac{b^2}{a^2} \\ &= -\frac{\sqrt{a^2-b^2}}{a}. \end{align*}

The equation of the tangent to the ellipse through $P$ therefore has the equation $y-\frac{b^2}{a}=-\frac{\sqrt{a^2-b^2}}{a}\left(x-\sqrt{a^2-b^2}\right).$

We can use this equation to work out the coordinates of $Q$.

When $x=0$ we get point $Q$, where \begin{align*} y &= \frac{b^2}{a}-\frac{\sqrt{a^2-b^2}}{a}\left(-\sqrt{a^2-b^2}\right)\\ &= \frac{b^2+(a^2-b^2)}{a}=a, \end{align*}

so $Q$ has coordinates $(0,a)$. (Without loss of generality we are assuming $a>b>0$.)

Now let’s work out the equation of the normal at $P$ and the coordinates of $R$.

Similarly, the equation of the normal at $P$ is $y-\frac{b^2}{a}=\frac{a}{\sqrt{a^2-b^2}}\left(x-\sqrt{a^2-b^2}\right).$ When $x=0$ we get the point $R$, where \begin{align*} y&=\frac{b^2}{a}+\frac{a}{\sqrt{a^2-b^2}}\left(-\sqrt{a^2-b^2}\right)\\ y&=\frac{b^2}{a}-a=\frac{b^2-a^2}{a}, \end{align*}

so $R$ has coordinates $\left(0,\dfrac{b^2-a^2}{a}\right)$.

Now we want to prove that $QR = HP$.

The other focus of the ellipse, $H$, has coordinates $\left(-\sqrt{a^2-b^2},0\right)$.

Therefore we have \begin{align*} HP &= \sqrt{(2\sqrt{a^2-b^2})^2+\left(\frac{b^2}{a}\right)^2} \\ &= \sqrt{4(a^2-b^2)+\frac{b^4}{a^2}}\\ &= \frac{1}{a}\sqrt{4a^4-4a^2b^2+b^4} \\ &= \frac{1}{a}\sqrt{(2a^2-b^2)^2}=\frac{2a^2-b^2}{a}. \end{align*}

Similarly, $QR = a+\frac{a^2-b^2}{a}=\frac{2a^2-b^2}{a}$ since we know that $Q$ is above the $x$-axis and $R$ is below it (had we chosen the other focus as $S$ this would be reversed.)

Thus $QR = HP$ as required.

For an alternative solution using the geometry of the ellipse, look at Solution 2.