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Question

Find the area of a triangle formed by the lines joining the vertex of the parabola x2=12y to the ends of its latus rectum.


Solution

for diagram upward parabola is along y-axis & o is the origin at (0,0) & on left side the point is L' (-6,3) & on right side point L(6,3) .........

THE GIVEN PARABOLA IS X2=12y

4a= 12

a= 3

therefore co-ordinates of O,L& L' are (0,0) (6,3) & (-6,3)

then ar(triangle O L L')

apply  1/2 I (x1y2 - x2y1) + (x2y3 -x3y2) + ( x3y1 - x1y3) I

= 1/2 I 0 -0 + 18 +18 - 0 - 0 I

= 36/2 = 18 sq. unit

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