Let be two parabolas. If the area of the bounded region enclosed between is six times the area of the bounded region enclosed between the line and , then is equal to : [2026]

Question

Let $P_{1} : y = 4 x^{2} and 𝑃_{2} : 𝑦 = 𝑥^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1} and P_{2}$ is six times the area of the bounded region enclosed between the line $y = α x, α > 0$ and $P_{1}$ , then $α$ is equal to : [2026]

Smart Achievers · Accepted Answer

(2)

$Area bounded between P_{1} & P_{2} is$

$\int_{- 3}^{3} ((x^{2} + 27) - (4 x^{2})) d x$

$(P.O.I. of P_{1} & P_{2} is x = \pm 3)$

$= 2 \int_{0}^{3} (27 - 3 x^{2}) d x = 2 {[27 x - x^{3}]}_{0}^{3}$

$= 2 [81 - 27] = 108$

$∴$ $Area bounded between P_{1} & L is 18 sq. units$

$(Area between x^{2} = 4 a y & line x = m y is \frac{8 a^{2}}{3 m^{3}})$

$∴$ $Area between x^{2} = \frac{y}{4} & x = \frac{y}{α} is$

$\frac{8 {(\frac{1}{16})}^{2}}{3 {(\frac{1}{α})}^{3}} = 18$

$\Rightarrow \frac{\frac{8}{16 \cdot 16}}{\frac{3}{α^{3}}} = 18 \Rightarrow α^{3} = 2^{6} \cdot 3^{3}$

$\Rightarrow α = 12$

Solution

Q.
Let $P_{1} : y = 4 x^{2} and 𝑃_{2} : 𝑦 = 𝑥^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1} and P_{2}$ is six times the area of the bounded region enclosed between the line $y = α x, α > 0$ and $P_{1}$ , then $α$ is equal to : [2026]

1 15

2 12

3 8

4 6

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Solution

Q. Let P1:y=4x2 and 𝑃2:𝑦=𝑥2+27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y=αx, α>0 and P1, then α is equal to : [2026]

1 15

2 12

3 8

4 6

Want Us to Email you About Special Offers & Updates?

Q.
Let $P_{1} : y = 4 x^{2} and 𝑃_{2} : 𝑦 = 𝑥^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1} and P_{2}$ is six times the area of the bounded region enclosed between the line $y = α x, α > 0$ and $P_{1}$ , then $α$ is equal to : [2026]