The area enclosed by the parabolas x - -2y2 and x - 1 - 3y2 is ab- where a- b are co-prime- Then a-b is

Point of intersection of the parabolas, 2y^2 = 1 3y^2 ⇒ y = ± 1 ⇒ x = 2 So, the required area is, = 2 ∫_0^1 (1 3y^2) ( 2y^2) dy nbsp;= nbsp;2 ∫_0^1 ...

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Standard XII

Mathematics

Area between Two Curves

The area encl...

Question

The area enclosed by the parabolas $x = - 2 y^{2}$ and $x = 1 - 3 y^{2}$ is $\frac{a}{b},$ where $a, b$ are co-prime. Then $a + b$ is

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Solution

Point of intersection of the parabolas,
$- 2 y^{2} = 1 - 3 y^{2} \Rightarrow y = \pm 1 \Rightarrow x = - 2$

So, the required area is,
$= 2 1 \int 0 (1 - 3 y^{2}) - (- 2 y^{2}) d y = 2 1 \int 0 (1 - y^{2}) d y = 2 {[y - \frac{y^{3}}{3}]}_{0}^{1} = 2 \times \frac{2}{3} = \frac{4}{3}$

Hence, $a + b = 7$

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The area enclosed by the parabolas x=−2y2 and x=1−3y2 is ab, where a,b are co-prime. Then a+b is

Point of intersection of the parabolas, −2y2=1−3y2⇒y=±1⇒x=−2 So, the required area is, =21∫0(1−3y2)−(−2y2) dy=21∫0(1−y2) dy=2[y−y33]10=2×23=43 Hence, a+b=7

The area enclosed by the parabolas $x = - 2 y^{2}$ and $x = 1 - 3 y^{2}$ is $\frac{a}{b},$ where $a, b$ are co-prime. Then $a + b$ is