What can we say if a normal to an ellipse passes through this point?

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Implicitly differentiating the equation of the ellipse with respect to \(x\), we find \[\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0\] and therefore \[\begin{align*} \frac{dy}{dx} &= - \frac{b^2x}{a^2y} \\ &= - \frac{b^2 a \cos \theta}{a^2 b \sin \theta} \\ &= - \frac{b \cos \theta}{a \sin \theta}. \end{align*}\]

Since the normal is perpendicular to the tangent, it has slope \[m = -\frac{1}{dy/dx} = \frac{a \sin \theta}{b \cos \theta}.\]

Hence the equation of the line is \[y - b\sin\theta = \frac{a \sin \theta}{b \cos \theta}\left(x - a \cos \theta\right)\] which we can rearrange to obtain \[a^2 - b^2 = \frac{ax}{\cos \theta} - \frac{by}{\sin\theta}\] as required.

The minor axis is the shortest diameter of the ellipse (a diameter of the ellipse is a chord that passes through the centre).

The ends have coordinates \((0,\pm b)\) (assuming that \(a \geq b>0\), which we can say without loss of generality).

If we know there is a value of \(\theta\) between \(0\) and \(\pi/2\) such that the normal at \(P\) passes through one end of the minor axis, then from the diagram, this normal must pass through \((0, -b)\).

Substituting \((0,-b)\) into the equation of the normal gives \[a^2 - b^2 = \frac{b^2}{\sin\theta} \qquad \implies \qquad \left(\frac{a}{b}\right)^2 = 1 +\frac{1}{\sin\theta}.\]

But now since \(0 < \sin\theta < 1\) we conclude \[\left(\frac{a}{b}\right)^2 > 2 \qquad \implies \qquad \frac{b^2}{a^2} < \frac{1}{2}.\]

If an ellipse is a ‘squashed’ circle, then the eccentricity measures the degree of ‘squashedness’. We have the standard result that \(e = \sqrt{1-\dfrac{b^2}{a^2}}\).

In our case, the eccentricity \[ e = \sqrt{1-\frac{b^2}{a^2}} > \sqrt{1 - \frac 1 2} = \frac{1}{\sqrt{2}} \] as required.