What can we say if a point $P$ on an ellipse is directly above the focus?

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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 \(\left(\sqrt{a^2-b^2},\dfrac{b^2}{a}\right)\).

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 \[y-\frac{b^2}{a}=-\frac{\sqrt{a^2-b^2}}{a}\left(x-\sqrt{a^2-b^2}\right).\]

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 \(a>b>0\).)

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 \(\left(0,\dfrac{b^2-a^2}{a}\right)\).

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.