Area of the region, bounded by the parabolas $3 x^{2} = 16 y$ and $4 y^{2} = 9 x$ , is

4

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6

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8

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16

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Solution

The correct option is B $4$
Solving $3 x^{2} = 16 y, 4 y^{2} = 9 x$
$4 (\frac{3 x^{2}}{16})^{2} = 9 x$
$\frac{36 x^{4}}{256} = 9 x$
$x^{4} = 64 x$
$x (x^{3} - 43) = 0 => x = 0, 4$
Required area= Area $4 y^{2} = 9 x$ between $x = 0, 4$ -area $3 x^{2} = 16 y$ between $x = 0, 4$
$= 4 \int 0 \frac{3 x^{2}}{16}) d x = \frac{3}{2} . \frac{2}{3} x^{3 / 2} - \frac{x^{3}}{16}]_{0}^{4}$
$= x^{\frac{3}{2}} - \frac{x^{3}}{16}]_{0}^{4}$
$= 4 \sqrt{4} - 4$
$= 8 - 4 = 4 s q u n i t s$

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Similar questions

4 y^{2} = 9 x

3 x^{2} = 16 y

4 < x^{2} + y^{2} < 16

3 x^{2} - y^{2} > 0

d .

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y^{2} = 6 (x - 1)

y^{2} = 3 x

9 x^{2} + 16 y^{2} = 144

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