If the area bounded by the parabola y^2=16ax | vedclass

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(D) The equation of the parabola is $y^2=16ax$ and the line is $y=4mx$.Substituting $x = \frac{y^2}{16a}$ into the line equation $y = 4mx$,we get $y =

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$2$

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(D) The equation of the parabola is $y^2=16ax$ and the line is $y=4mx$.Substituting $x = \frac{y^2}{16a}$ into the line equation $y = 4mx$,we get $y = 4m(\frac{y^2}{16a}) = \frac{my^2}{4a}$.This gives $y^2 = \frac{4ay}{m}$,so $y(y - \frac{4a}{m}) = 0$. Thus,the intersection points are $y=0$ and $y=\frac{4a}{m}$.The area bounded by the curves is given by:$\int_0^{\frac{4a}{m}} (\frac{y}{4m} - \frac{y^2}{16a}) dy = \frac{a^2}{12}$Evaluating the integral:$[\frac{y^2}{8m} - \frac{y^3}{48a}]_0^{\frac{4a}{m}} = \frac{a^2}{12}$$\frac{(4a/m)^2}{8m} - \frac{(4a/m)^3}{48a} = \frac{a^2}{12}$$\frac{16a^2}{8m^3} - \frac{64a^3}{48am^3} = \frac{a^2}{12}$$\frac{2a^2}{m^3} - \frac{4a^2}{3m^3} = \frac{a^2}{12}$$\frac{6a^2 - 4a^2}{3m^3} = \frac{a^2}{12}$$\frac{2a^2}{3m^3} = \frac{a^2}{12}$$\frac{2}{3m^3} = \frac{1}{12}$$m^3 = \frac{2 	imes 12}{3} = 8$$m = 2$Thus,option $(D)$ is correct.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

If the area bounded by the parabola $y^2=16ax$ and the line $y=4mx$ is $\frac{a^2}{12}$ sq. units,then the value of '$m$' is

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