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

Step 1 :
The given equation of the parabola is
x2=12yx2=12y
Now comparing this with the general equation x2=4ayx2=4ay we get,
4a=12⇒a=34a=12⇒a=3
∴∴ The coordinates of foci are F(0,a)F(0,a). Let AB be the latus rectum of the given parabola.

Step 2
At y=3,x2=12(3)y=3,x2=12(3)
⇒x2=36⇒x2=36
∴x=±6∴x=±6
∴∴ The coordinates of A are (-6, 3) and coordinates of B are (6,3)
∴∴ The vertices of Δ0ABΔ0AB are
0(0,0),A(−6,3)andB(6,3)0(0,0),A(−6,3)andB(6,3)
Step 3 :
Area of the triangle is 1212∣∣∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣∣∣|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|
Area of a triangle ABC, whose coordinates are (x1,y1),(x2,y2),(x3,y3)(x1,y1),(x2,y2),(x3,y3)
Substituting the values we get,
area of Δ0ABΔ0AB
1212∣∣∣0(3−3)+(−6)(3−0)+6(0−3)∣∣∣|0(3−3)+(−6)(3−0)+6(0−3)| sq.units
=12=12∣∣∣−36−36∣∣∣|−36−36| sq.units
=12=12∣∣∣−36∣∣∣|−36| sq.units
=12=12×36×36 sq.units
= 18 sq.units
Hence the required area of the triangle is 18 sq.units.