COMEDK 2025 Evening Solved Paper

Concept:The area of a triangle formed by the vertex and the ends of the latus rectum of a parabola is calculated using the coordinates of these points.Explanation:The given parabola is $x^2 = \lambda y$.Compare it with the standard form $x^2 = 4ay$.This gives $4a = \lambda$, so $a = \frac{\lambda}{4}$.Vertex is at $(0,0)$.Focus is at $(0,a) = (0, \frac{\lambda}{4})$.Endpoints of the latus rectum are $(\pm 2a, a) = (\pm \frac{\lambda}{2}, \frac{\lambda}{4})$.Base of triangle = distance between endpoints = $\lambda$.Height of triangle = vertical distance from vertex $(0,0)$ to the line $y = \frac{\lambda}{4}$ = $\frac{\lambda}{4}$.Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \lambda \times \frac{\lambda}{4} = \frac{\lambda^2}{8}$.Given area = 18 sq units, so $\frac{\lambda^2}{8} = 18$.Multiply both sides by 8: $\lambda^2 = 144$.Taking positive square root (since $\lambda$ is positive for standard parabola), $\lambda = 12$.Answer:$\lambda = 12$