Question:
The area of the triangle formed by the lines joining the vertex of the parabola \(x^2 = 48y\) to the ends of its latus rectum, is .....
The area of the triangle formed by the lines joining the vertex of the parabola \(x^2 = 48y\) to the ends of its latus rectum, is .....
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Find a from 4a = 48, then the ends of the latus rectum are (plus or minus 2a, a), and use the triangle area.
Updated On: Oct 1, 2026
- \(144\) sq. units
- \(288\) sq. units
- \(72\) sq. units
- \(576\) sq. units
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For a parabola \(x^2 = 4ay\), the vertex is the origin, the focus is \((0, a)\) and the latus rectum is the chord through the focus at \(y = a\). Its length is \(4a\).
Step 2: Detailed Explanation:
Compare \(x^2 = 48y\) with \(x^2 = 4ay\): \(4a = 48\), so \(a = 12\).
The ends of the latus rectum are \((\pm 2a, a) = (\pm 24, 12)\).
Step 3: Find the area.
The triangle has the vertex \((0, 0)\) and the two ends \((24, 12)\) and \((-24, 12)\). Its base is \(48\) (the latus rectum) and its height is \(12\).
\[ \text{Area} = \frac{1}{2}\times 48 \times 12 = 288\text{ sq. units} \]
Final Answer:
The area is 288 square units, option (B). \[ \boxed{288\text{ sq. units}} \]
For a parabola \(x^2 = 4ay\), the vertex is the origin, the focus is \((0, a)\) and the latus rectum is the chord through the focus at \(y = a\). Its length is \(4a\).
Step 2: Detailed Explanation:
Compare \(x^2 = 48y\) with \(x^2 = 4ay\): \(4a = 48\), so \(a = 12\).
The ends of the latus rectum are \((\pm 2a, a) = (\pm 24, 12)\).
Step 3: Find the area.
The triangle has the vertex \((0, 0)\) and the two ends \((24, 12)\) and \((-24, 12)\). Its base is \(48\) (the latus rectum) and its height is \(12\).
\[ \text{Area} = \frac{1}{2}\times 48 \times 12 = 288\text{ sq. units} \]
Final Answer:
The area is 288 square units, option (B). \[ \boxed{288\text{ sq. units}} \]
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