Question

The base of a triay is x+y-2=0 and opposite vertex is (2,-1). Find the equatios if the remaining sides.

Answer 1

Step-by-step explanation:

The given parabola is x

2

=12y

On comparing this equation with x

2

=4ay, we get

⇒4a=12

⇒a=3

⇒So, the coordinates of focus is S(0,a)=S(0,3)

⇒Let AB be the latus rectum of the given parabola.

⇒Coordinates of end-points of latus rectum are (−2a,a),(2a,a)

⇒Coordinates of A are (−6,3) while the coordinates of B are (6,3)

Therefore, the vertices of are △OAB are O(0,0),A(−6,3) and B(6,3)

⇒Let (x

1

,y

1

),(x

2

,y

2

),(x

3

,y

3

) be the coordinates of ΔOAB

⇒Area of △ OAB

=

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

2

)+x

3

(y

1

−y

2

)∣

=

2

1

∣0(3−3)+(−6)(3−0)+6(0−3)∣

=

2

1

∣(−6)(3)+6(−3)∣

=

2

1

∣−18−18∣

=

2

1

∣−36∣

=

2

1

×36

=18 sq.units

Thus the required area of the triangle is 18 unit

2