The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2y^2=2\Longleftrightarrow a^2\cos^2{t}+2b^2\sin^2{t}=2.$$
Trig identities gives 
$$a^2(1-\sin^2{t})+2b^2\sin^2{t}=a^2-a^2\sin^2{t}+2b^2\sin^2{t}=a^2-\sin^2{t}(a^2+2b^2)=2.$$
Solving for $\sin{t}$ gives
$$\sin{t}=\pm\sqrt{\frac{a^2-2}{a^2+2b^2}}=\pm f(a,b).$$
Since $$-1\leq\sin{t}\leq1\Longleftrightarrow -1\leq\pm f(a,b)\leq1.$$
Answer a) gives $f(2,1)=\frac{\sqrt{3}}{3} \in[-1,1]$. OK!
Answer b) gives $f(1,2)=\frac{i}{3}\in \mathbb{C}$. Disregard.
Answer c) gives $f(1,\sqrt{2}) = \frac{i\sqrt{5}}{5}\in \mathbb{C}$. Disregard.
So according to me, the correct answer should be a). But correct answer is d). Where did I go wrong?

For Parseval...
\begin{align}
x^2+2y^2 &= 2 \\
a²\cos^2(t) + 2b^2 \sin^2(t) &= 2 \\
\cos^2(t) + \frac {2b^2}{a^2} \sin^2(t) &= \frac 2 {a^2} \\
1-\sin^2(t) + \frac {2b^2}{a^2} \sin^2(t) &= \frac 2 {a^2} \\
(\frac {2b^2}{a^2} -1)\sin^2(t) &= \frac 2 {a^2} -1 \\
( {2b^2}- a^2 )\sin^2(t) &=  2 - {a^2} 
\end{align}
This must be true for any t...You can't solve it for t.
So you need to have:
$$0*\sin^2(t) = 0$$
Therefore, 
\begin{cases}
(2b^2- a^2 )=0 \iff b²= \frac {a^2}{2}=1 \\
2 - {a^2}=0  \iff a=\sqrt 2
\end{cases}
The solution f(2,1) only works for a certain t (=$\pi$ /2) in your formula Parseval...
$( 2- 4 )\sin^2(t)=  2 - 4$
$2\sin^2(t)=  2$
$\sin^2(t)=  1$
It's not true therefore for any t in [0,$2\pi$]....
