If the area enclosed by the parabolas P 1 : 2 y = 5 x 2 and P 2 : x 2 - y + 6 = 0 is equal to the area enclosed by P 1 and y = x , 0 , then 3 is equal to _______ . [2023]

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Solution

Q.
If the area enclosed by the parabolas $P_{1} : 2 y = 5 x^{2}$ and $P_{2} : x^{2} - y + 6 = 0$ is equal to the area enclosed by $P_{1}$ and $y = α x, α > 0$ , then $α^{3}$ is equal to _______ . [2023]

Ans.
(600)

Given, $P_{1} : 2 y = 5 x^{2}, P_{2} : x^{2} - y + 6 = 0 and y = α x$

The point of intersection of $P_{1}$ and $P_{2}$ is given by

$2 y = 5 (y - 6) \Rightarrow 2 y = 5 y - 30 \Rightarrow y = 10$

Put $y = 10$ in $P_{1}$ , we get $2 (10) = 5 x^{2}$

$\Rightarrow 5 x^{2} = 20 \Rightarrow x = \pm 2$

Thus $P_{1}$ and $P_{2}$ intersect at $(- 2, 10)$ and $(2, 10)$ .

$Area = 2 \int_{0}^{2} (x^{2} + 6 - \frac{5 x^{2}}{2}) d x$ $= 2 {[\frac{x^{3}}{3} + 6 x - \frac{5 x^{3}}{6}]}_{0}^{2}$

$= 2 [(\frac{8}{3} + 12 - 5 \times \frac{8}{6}) - 0] = 2 (8) = 16 sq. units$

Now, the point of intersection of $2 y = 5 x^{2}$ and $y = α x$ , is given by

$\Rightarrow 2 α x = 5 x^{2} \Rightarrow x = \frac{2 α}{5}$

$Area = \int_{0}^{2 α / 5} (α x - \frac{5 x^{2}}{2}) d x = {[\frac{α x^{2}}{2} - \frac{5 x^{3}}{6}]}_{0}^{2 α / 5}$

$= \frac{α}{2} (\frac{4 α^{2}}{25}) - \frac{5}{6} (\frac{8 α^{3}}{125}) = \frac{2 α^{3}}{25} - \frac{4 α^{3}}{75} = \frac{2 α^{3}}{75}$

$\frac{2 α^{3}}{75} = 16 [Given]$

$\Rightarrow α^{3} = 75 \times 8 = 600$

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