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