MHTCET 13 Apr 2026 Shift 1 Paper

Concept:For the parabola $x^2=4ay$, the latus rectum endpoints are $(\pm 2a, a)$, and the triangle with the vertex has area $\frac{1}{2} \times 4a \times a = 2a^2$.Explanation:Given parabola: $x^2 = 48y$.Compare with standard form $x^2 = 4ay$.So, $4a = 48$, hence $a = 12$.Vertex of the parabola is $(0,0)$.For $x^2 = 4ay$, the latus rectum is the line $y = a$.Thus, here $y = 12$.Put $y = 12$ in the parabola: $x^2 = 48(12) = 576$.So $x = \pm 24$.Ends of the latus rectum are $(-24, 12)$ and $(24, 12)$.The triangle is formed by these two points and the vertex $(0,0)$.Base of the triangle $= 24 - (-24) = 48$.Height is the perpendicular distance from $(0,0)$ to $y = 12$, which is $12$.Area $= \frac{1}{2} \times 48 \times 12 = 288$ sq. units.Answer:$288$ sq. units, i.e. Option B.