Question

`\underline{\boxed{\blue{\Large{\bf{Hello\:Brainlians}}}}}`

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 .



`\boxed{\sf{Note:-}}`


`\\ \tt\hookrightarrow Spams, plagarised\: from \:web , Short \:answers \:will \:be \:deleted\; on\:the\:spot.`


`\\ \tt\hookrightarrow Make\:sure\:you\:provide\:step\:by\:step\: solution\: without\: omitting\:any\:step.`

`\\ \tt\hookrightarrow Attach\:the\: Diagram\:for\: solution.`



`\boxed{\bf All\:the\:best!}`​

Answer 1

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.

Answer 2

18 sq. unit

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=
`=(1)/(2) *b*h\\\\=(1)/(2) [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)\\\\=(1)/(2) [0(3-3)+6(3-0)+(-6)(0-3)\\\\=(1)/(2) [18+18]\\\\\\=18.sq.units`