Parameterisation of an ellipse

Question

$u(x, y) := ϕ(x^2 + 4y^2)$.

For $t > 0$, let $E_{t}$ be the ellipse $\{(x, y) : x^2+ 4y^2= t\}.$

The ellipse has been parameterised to $r(τ ) = \sqrt{t}(\cos (τ), \frac{1}{2} \sin (τ) )$

Can you explain thisv please ?

Answer 1

For each value of $\tau$ the point $$ r(\tau) = \sqrt{t}(\cos \tau, (1/2)\sin \tau) = (x,y) $$ is on the ellipse because $$ x^2 + 4y^2 = t(\cos^2 \tau + \sin^2 \tau) = t . $$ When $\tau = 0$, $r(\tau)$ is the intersection of the ellipse and the positive $x$-axis. As $\tau$ increases to $2\pi$ it runs counterclockwise once around.