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Question
Find the area of the triangle formed by the lines joining the vertex of the parabola \[x^2 = 12y\] to the ends of its latus rectum.
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Solution
The given equation of the parabola is x2 = 12y.
On comparing the given equation with\[x^2 = 4ay\]
a = 3 
Required area = \[\frac{1}{2}\left( LL' \times OS \right) = \frac{1}{2} \times 12 \times 3 = 18 \text{ square units }\]
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