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

Hi there!

Parabola x² = 12y

→ x² = 4ay

→ 4a = 12

→ a = 12÷4

→ a = 3

So, the co-ordinates of the focus is:-

S(0,a)=(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 B's coordinates are (6,3).

→ ∆OAB are O(0,0), A(-6,3), B(6,3)

Area of ∆OAB is :-

(Solving part attached as image)

=> 18 unit² is the required answer.

18 sq. unit

Step-by-step explanation:

Parabola is x²=12y

On comparing it with

→x²=4ay

→4a=12

→a=12/4

→a=3

Finding the value of x.

→x²=12y

→x²=12×3

→x²=36

→x=√36

→x=6

When we have a=3 and x=6

(-x1,a)=(-6,3)

(x1,a)=(6,3)

Let,

(x1,y1)=(0,0)

(x2,y2)=(6,3)

(x3,y3)(-6,3)

Now,

Area=

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.

b

step-by-step explanation:

u take $75 minus $60.the anwer/original($75)multiply by 100