### Of ellipses, hyperbolae and mugging

For as long as I can remember, I have had unnatural inertia in studying coordinate geometry. It seemed to be a pursuit of rote learning and regurgitating requisite formulae, which is something I detested. My refusal to “mug up” formulae cost me heavily in my engineering entrance exams, and I was rather proud of myself for having stuck to my ideals in spite of not getting into the college of my dreams.

However, now I realise what useful entities ellipses and hyperbolae are in reality. Hence, as a symbolic gesture, I will derive the formulae of both the ellipse and the hyperbola in the most simple settings- that of the centre being at the origin $(0,0)$.

1. Ellipse- The sum of distances from two _foci_ is constant. Let the sum be “$L$“. As the centre is at the origin, and we are free to take the foci along the x-axis, the coordinates of the foci are $(-c,0)$ and $(c,0)$. We thus have the equation $\sqrt{(x-c)^2 +y^2}+\sqrt{(x+c)^2+ y^2}=L$. On simpifying this, we get $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a^2=\frac{L^2}{4}$ and $b^2=\frac{(L^2-4c^2)}{4}$.

2. In the case of a hyperbola, under similar conditions, we have the equation $\sqrt{(x-c)^2 +y^2}-\sqrt{(x+c)^2+ y^2}=L$. This under simplification gives $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a^2=\frac{L^2}{4}$ and $b^2=\frac{(4c^2-L^2)}{4}$.

### The utility of trigonometrical substitutions

Today we will discuss the power of trigonometrical substitutions.

Let us take the expression $\frac{\sum_{k=1}^{2499} \sqrt{10+\sqrt{50+\sqrt{k}}}}{\sum_{k=1}^{2499} \sqrt{10-\sqrt{50+\sqrt{k}}}}$

This is a math competition problem. One solution proceeds this way: let $p_k=\sqrt{50+\sqrt{k}}, q_k=\sqrt{50-\sqrt{k}}$. Then as $p_k^2+q_k^2=10^2$, we can write $p_k=10\cos x_k$ and $q_k=10\sin x_k$.

This is an elementary fact. But what is the reason for doing so?

Now we have $a_k=\sqrt{10+\sqrt{50+\sqrt{k}}}=\sqrt{10+10\cos x_k}=\sqrt{20}\cos \frac{x_k}{2}$. Similarly, $b_k=\sqrt{10-\sqrt{50+\sqrt{k}}}=\sqrt{10-10\cos x_k}=\sqrt{20}\sin \frac{x_k}{2}$. The rest of the solution can be seen here. It mainly uses identities of the form $2\sin A\cos B=(\sin A+\cos B)^2$ to remove the root sign.

What if we did not use trigonometric substitutions? What is the utility of this method?

We will refer to this solution, and try to determine whether we’d have been able to solve the problem, using the same steps, but not using trigonometrical substitutions.

$a_{2500-k}=\sqrt{10+\sqrt{50+\sqrt{2500-k}}}=\sqrt{10+\sqrt{50+\sqrt{(50+\sqrt{k})(50-\sqrt{k})}}}$

$=\sqrt{10+\sqrt{50+100\frac{\sqrt{50+\sqrt{k}}}{10}\frac{\sqrt{50-\sqrt{k}}}{10}}}=\sqrt{10+\sqrt{50}\sqrt{1+2\frac{\sqrt{50+\sqrt{k}}}{10}\frac{\sqrt{50-\sqrt{k}}}{10}}}$
$=\sqrt{10+10(\frac{\sqrt{50+\sqrt{k}}}{10\sqrt{2}}+\frac{\sqrt{50-\sqrt{k}}}{10\sqrt{2}})}=\sqrt{10+10\times 2\times\frac{\frac{1}{2}(\frac{\sqrt{50+\sqrt{k}}}{10\sqrt{2}}+\frac{\sqrt{50-\sqrt{k}}}{10\sqrt{2}})}{\sqrt{\frac{50+\sqrt{k}}{20\sqrt{2}}-\frac{50-\sqrt{k}}{20\sqrt{2}}+\frac{1}{2}}}\times \sqrt{\frac{50+\sqrt{k}}{20\sqrt{2}}-\frac{50-\sqrt{k}}{20\sqrt{2}}+\frac{1}{2}}}$

As one might see here, our main aim is to remove the square root radicals, and forming squares becomes much easier when you have trigonometrical expressions. Every trigonometrical expression has a counterpart in a complex algebraic expression. It is only out of sheer habit that we’re more comfortable with trigonometrical expressions and their properties.