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 \(QR=HP\).

Diagram of ellipse with the tangent and normal at P intersecting the y-axis at Q and R respectively

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.