$\mathrm C$onsider an ellipse with the origin as its centre, i.e., of the type $$\frac {x^2} {a^2} + \frac {y^2} {b^2} = 1$$ and a line joining two points on the ellipse.
$\mathrm T$he problem is to find the point, say, $\mathrm P (h, k)$, on the ellipse, furthest from this line.
I took the points as $ P(a \cos\theta, b \sin\theta), Q(a\cos\phi, b\sin\phi)$. To solve this, I tried to use the formula for perpendicular distance of a point from a line, which is $$\frac{|Ah + Bk + C|} {\sqrt {A^2 + B^2}}$$ For distance to be maximum, $Ah + Bk+ C$ must be maximum. Now, the equation to the line can be written in the form: $$y - b\sin\phi = \frac {b\sin\theta - b\sin\phi} {a\cos\theta - a\cos\phi} (x - a\cos\phi). $$ On simplifying, we can try to find the maximum of $Ah + Bk + C$. When this didn't work,I tried other, similar methods. I still haven't got anything useful, though.