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Find the area of the triangle formed by the lines joining the vertex of the parabola `x^2=12 y` to the ends of its latus-rectum.

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Equation of parabola : `x^(2)=12y`
Its axis is along y-axis.
Here, 4a=12
`rArr" "a=3`
`rArr" "OS=3`
Coordinates of `S-=(0,3)`
Let AS=h
`:.` Coordinates of `A-=(h,3)` This point lies on the parabola `x^(2)=12y`
`:." "h^(2)=12xx3=36`
`rArr" "h=pm6`
Now, `AS = 6" "rArr" "AB=2AS=12`
Area of `DeltaAOB=(1)/(2)xxABxxOS`
`(1)/(2)xx12xx3=18` sq. units
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