Strangely worded, but I think they want you to express y as a function of x. For example in the first one if you use the identity $\displaystyle \sin \theta = \sqrt {1- \cos^2 \theta}$ you can get an expression for y(x) which has no $\displaystyle \theta$ term in it.
Trigonometric identities needed for these 2 questions:
(a) cos^2(theta) + sin^2(theta) = 1. With that in mind, cos(theta) = x/a and sin(theta) = y/b. That's how you have (x/a)^2+(y/b)^2 = 1
(a) tan^2(theta) + 1 = sec^2(theta). With that in mind, tan(theta) = x/a and sec(theta) = y/b. Then you have (x/a)^2 + 1 = (y/b)^2.
Good day.
Did you try the approach I suggested? It gets to the same answer:
$\displaystyle y = b \sin \theta$
$\displaystyle y = b \sqrt {1- \cos^2 \theta} $
$\displaystyle y^2 = b^2(1- \cos^2 \theta) $
$\displaystyle y^2 = b^2(1 - (\frac x a)^2) $
$\displaystyle \frac {y^2}{b^2} + \frac {x^2}{a^2} = 1 $