Class CBSE Class 12 Mathematics Applications of Integrals Q #929
COMPETENCY BASED
APPLY
5 Marks 2023 LA
33. Using integration, find the area of the region bounded by the parabola $y^{2}=4ax$ and its latus rectum.

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Step-by-Step Solution

The equation of the parabola is $y^2 = 4ax$. The latus rectum is $x = a$.

The area bounded by the parabola and the latus rectum can be found by integrating the function $x = \frac{y^2}{4a}$ with respect to $y$ from $-2a$ to $2a$. Due to symmetry, we can integrate from $0$ to $2a$ and multiply by 2.

Area = $2 \int_{0}^{2a} (a - \frac{y^2}{4a}) dy$

Area = $2 [ay - \frac{y^3}{12a}]_{0}^{2a}$

Area = $2 [(a(2a) - \frac{(2a)^3}{12a}) - (0 - 0)]$

Area = $2 [2a^2 - \frac{8a^3}{12a}]$

Area = $2 [2a^2 - \frac{2a^2}{3}]$

Area = $2 [\frac{6a^2 - 2a^2}{3}]$

Area = $2 [\frac{4a^2}{3}]$

Area = $\frac{8a^2}{3}$

Correct Answer: $\frac{8a^2}{3}$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concept of integration to find the area bounded by a parabola and its latus rectum. This requires using a known formula and applying it to a specific geometric situation.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to execute a specific procedure (integration) to arrive at the solution. It involves knowing the steps to set up the integral and then performing the integration.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of integration to solve a geometric problem, which aligns with competency-based assessment.