Conics: ellipse in standard position

An example with answer I have goes;
x = 6 cos t
y = sqrt 11 sin t

Write down the equation of the conic & state what type of conic it is.

trig identity cos²t + sin²t = 1

which gives

x² / 6² + y² / (sqrt 11)² = 1 . . . . . . . .

er! how does this happen, if you could help!

$\displaystyle x = 6\cos(t), \ y = \sqrt{11}\sin(t)$, so

$\displaystyle \cos(t) = \frac{x}{6}, \ \sin(t) = \frac{y}{\sqrt{11}}$, so

$\displaystyle \cos^2(t) = \left(\frac{x}{6}\right)^2, \ \sin^2(t) = \left(\frac{y}{\sqrt{11}}\right)^2$, so

$\displaystyle \cos^2(t) = \frac{x^2}{6^2}, \ \sin^2(t) = \frac{y^2}{(\sqrt{11})^2}$.

NOW, use the trigonometry formula $\displaystyle \cos^2(t) + \sin^2(t) = 1$, and plug in what we just found.

$\displaystyle \cos^2(t) + \sin^2(t) = 1$, so

$\displaystyle \left( \frac{x^2}{6^2}\right) + \left( \frac{y^2}{(\sqrt{11})^2} \right) = 1$, so

$\displaystyle \frac{x^2}{6^2} + \frac{y^2}{(\sqrt{11})^2} = 1$.

You could rewrite that as:

$\displaystyle \frac{x^2}{36} + \frac{y^2}{11} = 1$.

Did that help?

A simple answer for a simple problem - thank you for taking the time. :-)