The area of the region bounded by the parabol | vedclass

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(A) The given equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,which implies $a = 3$.The focus of the parabol

Vedclass · Accepted Answer

$24$

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(A) The given equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,which implies $a = 3$.The focus of the parabola is $(0, a) = (0, 3)$.The equation of the latus rectum is $y = 3$.The region is bounded by the parabola $y = \frac{x^2}{12}$ and the line $y = 3$.The points of intersection are found by substituting $y = 3$ into $x^2 = 12y$,giving $x^2 = 36$,so $x = \pm 6$.The area $A$ is given by the integral $A = \int_{-6}^{6} (3 - \frac{x^2}{12}) dx$.Since the function is even,$A = 2 \int_{0}^{6} (3 - \frac{x^2}{12}) dx$.$A = 2 [3x - \frac{x^3}{36}]_{0}^{6} = 2 [3(6) - \frac{216}{36}] = 2 [18 - 6] = 2(12) = 24$ sq. units.

The area of the region bounded by the parabola $x^2 = 12y$ and its latus rectum is . . . . . . sq. units.

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