Consider the parabola x^2 = 12y . Which of the following statements a…

The correct answer is Option (3) → (A) and (B) only

Comparing with \(x^{2} = 4ay\) gives a = 3, so the focus is (0, 3) and the latus rectum lies along y = 3. It meets the parabola where \(x^{2} = 36\), that is at x = ±6.

Between x = −6 and x = 6 the line y = 3 is above the curve \(y = \frac{x^{2}}{12}\):

    \( \displaystyle \int_{-6}^{6}\left(3-\frac{x^{2}}{12}\right)dx = \left[3x-\frac{x^{3}}{36}\right]_{-6}^{6} \)

    \( = (18-6)-(-18+6) = 24 \)

The parabola is symmetric about the y-axis, so the first-quadrant part is half of this:

    \( \displaystyle \frac{24}{2} = 12 \)

The line y = 12 meets the parabola where \(x^{2} = 144\), that is at x = ±12:

    \( \displaystyle \int_{-12}^{12}\left(12-\frac{x^{2}}{12}\right)dx = \left[12x-\frac{x^{3}}{36}\right]_{-12}^{12} \)

    \( = (144-48)-(-144+48) = 192 \)

So the latus rectum is y = 3, the area it cuts off is 24, the first-quadrant part is 12, and the area cut off by y = 12 is 192.

Checking the statements

(A) The latus rectum of the parabola lies along the line y = 3 — Correct. a = 3, so the focus is (0, 3).

(B) The area bounded by the parabola and its latus rectum is 24 sq. units — Correct. The integral from −6 to 6 is 24.

(C) The first-quadrant area bounded by the parabola, its latus rectum and the y-axis is 6 sq. units — Incorrect. It is half of 24, which is 12.

(D) The area bounded by the parabola and the line y = 12 is 96 sq. units — Incorrect. The integral from −12 to 12 is 192.