Question

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

Answer 1

The given equation of the parabola isx2=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.


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 are0(0,0),A(−6,3)and B(6,3)

Area of the triangle is 1/2[x1(y2-y2)+x2(y3-y1)+x2(y1-y2)]
Area of a triangle ABC, whose coordinates are (x1,y1),(x2,y2),(x3,y3)

Substituting the values we get,area of Δ0AB
=1/2[0(3-3)+(-6)(3-0)+6(0-3)
=1/2[-18-18]
=1/2[-36]
=1/2*-36
= 18 sq.unitsHence the required area of the triangle is 18 sq.units.

Answer 2

x² = 12y
we know general equation is x² = 4ay then vertex is ( 0, 0) and latus reactum is y = a

compare , x² = 12y = 4(3)y
and x² = 4ay
then,
a = 3
so, eqn of Latus rectum ; y = 3
put y = 3 in eqn of parabola , x² = 12y

x² = 36
x = ± 6
hence, ( 6, 3) and ( -6 ,3) are the points lies on line and latus rectum .

now , a/ c to question ,
area of ∆ is formed by ( 0, 0) , ( 6, 3) and ( - 6, 3)

use co- ordinate formula for finding area of ∆ .
area of ∆ = 1/2{ 0( 3 - 3) + 6( 3 - 0) - 6( 0-3) }

= 1/2{ 0 + 18 + 18 }

= 18 square unit

hence, area of ∆ = 18 square unit .