Explanation

(b)
Putting $x = {at}^{2}$ in $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,
We get, $t^{4} + \frac{y^{2}}{b^{2}} = 1$
$ie$, $y^{2} = b^{2}\left( 1 - t^{4} \right) = b^{2}\left( 1 + t^{2} \right)(1 - t^{2})$
$y$ is real, if $1 - t^{2} \geq 0$
$ie,\ |t| \leq 1$