Answer
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$; Ellipse equation
Work Step by Step
Re-arrange the parametric equations as shown below:
$\cos t=\dfrac{x-h}{a}$ and $\sin t=\dfrac{y-k}{b}$
Since, $\sin^2 t+\cos^2 t=1$
$\implies (\dfrac{y-k}{b})^2 + (\dfrac{x-h}{a})^2=1$.
or, $\dfrac{(y-k)^2}{b^2}+\dfrac{(x-h)^2}{a^2}=1`$
Therefore, $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
This is the ellipse equation
