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Quadrilaterals

If that the a...

Question

If that the area bounded between parabolas $y = 2 x^{2}$ and $y = x^{2} + 4$ is 32/a sq. units. Find $a$ .

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Solution

$y = 2 x^{2}$ , $y^{2} = x^{2} + 4$
$A = (2, 8)$ , $B = (- 2, 8)$ , $C = (0, 0)$
Area of triangle ABCA
$= \int_{0}^{2} (x^{2} + 4 - 2 x^{2}) d x$
$\frac{16}{3}$
Due to symmetry
$A r e a = \frac{32}{3}$
$a = 3$

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