The area of the region common to the parabola | vedclass

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(B) Step $1$: Rewrite the equations as $y^2 = \frac{9}{4}x$ and $x^2 = \frac{16}{3}y$.Step $2$: Find the intersection points by substituting $y = \fra

Vedclass · Accepted Answer

$4 	ext{ sq. units}$

Vedclass · Answer

(B) Step $1$: Rewrite the equations as $y^2 = \frac{9}{4}x$ and $x^2 = \frac{16}{3}y$.Step $2$: Find the intersection points by substituting $y = \frac{x^2}{16/3} = \frac{3x^2}{16}$ into the first equation: $(\frac{3x^2}{16})^2 = \frac{9}{4}x \implies \frac{9x^4}{256} = \frac{9}{4}x \implies x^4 = 64x$.Step $3$: Solving $x(x^3 - 64) = 0$ gives $x = 0$ and $x = 4$. The corresponding $y$ values are $0$ and $3$.Step $4$: The area $A$ is given by $\int_{0}^{4} (\sqrt{\frac{9}{4}x} - \frac{3x^2}{16}) dx$.Step $5$: $A = \int_{0}^{4} (\frac{3}{2}x^{1/2} - \frac{3}{16}x^2) dx = [\frac{3}{2} \cdot \frac{2}{3}x^{3/2} - \frac{3}{16} \cdot \frac{x^3}{3}]_{0}^{4} = [x^{3/2} - \frac{x^3}{16}]_{0}^{4}$.Step $6$: $A = (4^{3/2} - \frac{4^3}{16}) = (8 - 4) = 4 	ext{ sq. units}$.

The area of the region common to the parabolas $4y^2 = 9x$ and $3x^2 = 16y$ is...

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