Prove that the equation of the normal to the ellipse \(x^2/a^2+y^2/b^2=1\) at the point \(P\,(a\cos\theta,b\sin\theta)\) is \[\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2-b^2.\]

If \(x = f(t), y = g(t)\), what is \(\dfrac{dy}{dx}\) in terms of \(t\)?

Or could we use implicit differentiation?

If 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, show that the eccentricity of the ellipse must be greater than \(1/\sqrt{2}\).

You might use this applet to explore the problem.

The eccentricity, \(e\), is a measure of how “squashed” the ellipse is. It can be defined in terms of the major and minor axes. \[e = \sqrt{1-\dfrac{b^2}{a^2}}\]