Question

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

Answer 1

The problem can be represented as follows, where ABAB is the latus rectum of the parabola x2=12yx2=12y.


Given x2=12y→4a=12→a=3x2=12y→4a=12→a=3.

Therefore, the coordinates of point C=(0,3)C=(0,3)


When y=3→x2=12×3=36→x=±6y=3→x2=12×3=36→x=±6

Therefore A=(−6,3)A=(−6,3) and B=(6,3)B=(6,3).

Given the 3 vertices of the triangle, we can calculate the area as follows: Area=|12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|Area=|12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|

Now that we know the vertices of ΔAOBΔAOB, we can calculate the area.=|12(0(3−3)+(−6)(3−0)+6(0−3)|=|12(0(3−3)+(−6)(3−0)+6(0−3)|=|12(−18−18)|=|12(−18−18)|=12=12×36=18sq. units