Review question

# Where does this chord of an ellipse cut the $x$-axis? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6843

## Solution

Prove that the equation of the chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ joining the points $(a\cos\alpha,b\sin\alpha)$ and $(a\cos\beta,b\sin\beta)$ is $\frac{x}{a}\cos\left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b}\sin\left(\frac{\alpha+\beta}{2}\right)= \cos\left(\frac{\alpha-\beta}{2}\right).$

The following standard identities will be useful to us:

$\sin x- \sin y = \sin \dfrac{x-y}{2}\cos \dfrac{x+y}{2}, \cos x -\cos y = -2 \sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2}.$

The gradient of the chord $HK$ is $\frac{b\sin\alpha-b\sin\beta}{a\cos\alpha-a\cos\beta}$ $=-\dfrac{b}{a} \frac{\sin \dfrac{x-y}{2}\cos \dfrac{x+y}{2}}{\sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2}}$ $=-\dfrac{b}{a} \frac{\cos \dfrac{x+y}{2}}{\sin \dfrac{x+y}{2}}.$

Therefore the equation of the line is $y-b\sin\alpha=-\dfrac{b}{a} \frac{\cos \frac{\alpha+\beta}{2}}{\sin \frac{\alpha+\beta}{2}}(x-a\cos\alpha)$ or $xb\cos\left(\frac{\alpha+\beta}{2}\right)+ay\sin\left(\frac{\alpha+\beta}{2}\right)= ab \sin \frac{\alpha + \beta}{2}\sin \alpha + ab \cos \frac{\alpha + \beta}{2}\cos \alpha,$ which simplifies to $\frac{x}{a}\cos\left(\frac{\alpha+\beta}{2}\right)+\frac{y}{b}\sin\left(\frac{\alpha+\beta}{2}\right)= \cos\left(\frac{\alpha-\beta}{2}\right).$

as required.

We can find the equation of the tangent through $H$ by allowing $\beta$ to tend to $\alpha$, giving us $bx\cos\alpha+a\sin\alpha-ab=0.$

Through a point $P$ on the major axis of an ellipse a chord $HK$ is drawn. Prove that the tangents at $H$ and $K$ meet the line through $P$ at right angles to the major axis at points equidistant from $P$.

We can find the point $P(x_p,0)$ by substituting $y=0$ into the equation of the chord found above: $x_p=a\frac{\cos\left(\frac{\alpha-\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)}.$

From our work above, the tangent through $H$ is $bx\cos\alpha+a\sin\alpha-ab=0,$ while the equation of the tangent though $K$ is $bx\cos\beta+a\sin\beta-ab=0.$

We are looking to prove that $PH’ =PK’$, where $H'$ and $K'$ are the intersections of the tangents with the line through $P$.

Let $H'$ be $(x_p,y_h)$ and $K'$ be $(x_p,y_k)$. The signs of $y_h$ and $y_k$ will always be opposite, since the chord $HK$ crosses the $x$-axis.

Consider the equation $(bx\cos\alpha+ay\sin\alpha-ab)(bx\cos\beta+ay\sin\beta-ab)=0.$

This must represent the equation of the pair of straight lines given by the tangent at $H$ and the tangent at $K$.

Where does the line $x = x_p$ cut these lines? At $H'$ and $K'$. So let’s substitute $x_p$ for $x$ in this equation, giving

$(bx_p\cos\alpha+ay\sin\alpha-ab)(bx_p\cos\beta+ay\sin\beta-ab)=0.$

This is a quadratic in $y$, say $Ay^2+By+C=0$. Its roots must be $y_h$ and $y_k$, and the sum of these roots is $-\dfrac{B}{A}$. Thus

$y_h+y_k = -\dfrac{a\sin\beta(bx_p\cos\alpha-ab)+a\sin\alpha(bx_p\cos\beta-ab)}{a^2\sin\alpha\sin\beta}$ $=-b\dfrac{x_p\sin(\alpha+\beta)-a(\sin\alpha+\sin\beta)}{a\sin\alpha\sin\beta}$ $=-b\dfrac{\dfrac{\cos\dfrac{\alpha-\beta}{2}}{\cos\dfrac{\alpha + \beta}{2}}2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha+\beta}{2}-(\sin\alpha+\sin\beta)}{\sin\alpha\sin\beta}$ $=-b\dfrac{2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2}-(\sin\alpha+\sin\beta)}{\sin\alpha\sin\beta}$

The following standard identity is now useful: $2\sin x\cos y = \sin (x+y)+\sin(x-y) \implies 2\sin\dfrac{\alpha+\beta}{2}\cos\dfrac{\alpha-\beta}{2} = \sin\alpha+\sin\beta.$

And so we have that $y_h+y_k =0$, and the lengths $PH'$ and $PK'$ are equal, as required.