Question

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

Answer 1

given equation of the parabola is

x2=12y

Now comparing this with the general equation x2=4ay we get,

4a=12⇒a=3

∴ The coordinates of foci are F(0,a). Let AB be the latus rectum of the given parabola.

Step 2

At y=3,x2=12(3)

⇒x2=36

∴x=±6

∴ The coordinates of A are (-6, 3) and coordinates of B are (6,3)

∴ The vertices of Δ0AB are

0(0,0),A(−6,3)andB(6,3)

Step 3 :

Area of the triangle is 1/2∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣

Area of a triangle ABC, whose coordinates are (x1,y1),(x2,y2),(x3,y3)

Substituting the values we get,

area of ΔOAB

1/2∣0(3−3)+(−6)(3−0)+6(0−3)∣ sq.units

=1/2∣−36−36∣ sq.units

=1/2∣−36∣ sq.units

=1/2×36 sq.units

= 18 sq.units

Hence the required area of the triangle is 18 sq.units.