<s>[INST] The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 11th student.
My work is below.
$\implies y = \frac{x^2}2 \\
\implies 2y = x^2 \\
\implies -x^2 +2y = 0 \ \ \ \  (1)$ 
and 
$x^2 + y^2 = 8 \ \ \ \  (2) $
Adding (1) and  (2)
$y^2 + 2y = 8 \\
(y-2)(y+4) =0 \\
\therefore y = 2 or -4$
But when I plot it in the plain with help of https://www.desmos.com/ value of y can only be 2 for the point of intersections. 

I'm unable to understand how I can eliminate -4 from the results.
 [/INST] I can substitute: $(x^2/2)^2+x^2=8$. I obtain: $x^4+4x^2-32=0$. Let $t=x^2$: $t^2+4t-32=0$. In other words: $(t+8)(t-4)$. $x^2=-8$ is impossible; $x^2=4$ has two solutions: $x=2$ or $x=-2$
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