The area inside the parabola 5x² - y = 0 but outside the parabola 2x² - y + 9 = 0 is:

$12\sqrt{3}$

Eliminating y, we get : 54x² - (2x² + 9) = 0 or 3x² = 9 $x = -\sqrt{3}, 3$

The two parabolas are $x^2=\frac{1}{2}y,\,x^2=\frac{1}{2}(y-9)$ and clearly from the figure second parabola is upper and first is lower.

∴ Required area = $2\int_0^{\sqrt{3}}(y_1-y_2)dx$ ... (i)

$=2\int\limits_0^{\sqrt{3}}[(2x^2+9)-5x^2].dx=2\int\limits_0^{\sqrt{3}}(9-3x^2).dx=12\sqrt{3}$