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(B) Given equations are $ay = x^2$ and $x + y = 2a$. From the first equation,$y = \frac{x^2}{a}$. Substituting this into the second equation: $x + \fr

Vedclass · Accepted Answer

$\frac{9}{2}$

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(B) Given equations are $ay = x^2$ and $x + y = 2a$. From the first equation,$y = \frac{x^2}{a}$. Substituting this into the second equation: $x + \frac{x^2}{a} = 2a$. Multiplying by $a$,we get $ax + x^2 = 2a^2$,which simplifies to $x^2 + ax - 2a^2 = 0$. Factoring the quadratic: $(x + 2a)(x - a) = 0$. Thus,the points of intersection are $x = -2a$ and $x = a$. The area $A$ is given by the integral of the upper curve minus the lower curve: $A = \int_{-2a}^{a} (2a - x - \frac{x^2}{a}) dx$. Integrating term by term: $A = [2ax - \frac{x^2}{2} - \frac{x^3}{3a}]_{-2a}^{a}$. Evaluating at the limits: $A = (2a^2 - \frac{a^2}{2} - \frac{a^3}{3a}) - (-4a^2 - \frac{4a^2}{2} - \frac{-8a^3}{3a})$. $A = (2a^2 - \frac{a^2}{2} - \frac{a^2}{3}) - (-4a^2 - 2a^2 + \frac{8a^2}{3})$. $A = (\frac{12a^2 - 3a^2 - 2a^2}{6}) - (\frac{-12a^2 - 6a^2 + 8a^2}{3})$. $A = \frac{7a^2}{6} - (\frac{-10a^2}{3}) = \frac{7a^2}{6} + \frac{20a^2}{6} = \frac{27a^2}{6} = \frac{9}{2}a^2$. Since the area is $ka^2$,we have $k = \frac{9}{2}$.

If the area of the region enclosed by the curve $ay = x^2$ and the line $x + y = 2a$ is $ka^2$,then $k =$

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