The maximum possible area bounded by the para | vedclass

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(B) Given the parabola $y=x^2+x+10$.Completing the square,we get $y = (x + \frac{1}{2})^2 + \frac{39}{4}$,or $(x + \frac{1}{2})^2 = (y - \frac{39}{4})

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$\frac{1}{6}$

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(B) Given the parabola $y=x^2+x+10$.Completing the square,we get $y = (x + \frac{1}{2})^2 + \frac{39}{4}$,or $(x + \frac{1}{2})^2 = (y - \frac{39}{4})$.This is a parabola of the form $X^2 = 4aY$ where $4a = 1$,so the length of the latus rectum is $1$.The area bounded by a parabola and a chord of length $L$ is given by the formula $A = \frac{1}{6} L^3 \cdot \frac{1}{a}$,where $4a$ is the length of the latus rectum.Here,$L = 1$ and $4a = 1$,so $a = \frac{1}{4}$.Substituting these values,the area $A = \frac{1}{6} 	imes (1)^3 	imes \frac{1}{1/4} = \frac{1}{6} 	imes 4 = \frac{2}{3}$ is incorrect based on standard formula application. Let's re-evaluate: The area between a parabola $y^2 = 4ax$ and a chord perpendicular to the axis is $\frac{2}{3} 	imes 	ext{base} 	imes 	ext{height}$. For a chord of length $L$ perpendicular to the axis,the height is $h = \frac{L^2}{8a}$.Here $L=1$ and $4a=1$,so $a=1/4$. Thus $h = \frac{1^2}{8(1/4)} = \frac{1}{2}$.Area $= \frac{2}{3} 	imes L 	imes h = \frac{2}{3} 	imes 1 	imes \frac{1}{2} = \frac{1}{3}$.Wait,the standard formula for the area cut off by a chord of length $L$ in a parabola with latus rectum $4a$ is $\frac{L^3}{6 \cdot (4a)} = \frac{1^3}{6 \cdot 1} = \frac{1}{6}$.

The maximum possible area bounded by the parabola $y=x^2+x+10$ and a chord of the parabola of length $1$ is

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