Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12 y to the ends of its latus rectum.

Solution:

The given parabola is x² = 12y.

On comparing this equation with x² = 4ay,

we obtain 4a = 12

⇒ a = 3

Therefore,

The coordinates of foci are ⇒ S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

 

At y = 3

⇒ x² = 12(3)

⇒ x² = 36

⇒ x = ± 6

Hence,

The coordinates of A are (- 6, 3) , while the coordinates of B are (6, 3)

Therefore, the vertices of ΔAOB are O (0, 0), A(- 6, 3) and B (6, 3)

ar (ΔAOB) = 1/2 |0(3 - 3) + (- 6)(3 - 0) + 6 (0 - 3)| unit²

= 1/2 |(- 6)(3) + 6 (- 3)| unit²

= 1/2 |- 18 - 18| unit²

= 1/2 |- 36| unit²

= 1/2 × 36 unit²

= 18 unit²

Thus, the required area of the triangle is 18 unit²

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NCERT Solutions Class 11 Maths Chapter 11 Exercise ME Question 6

Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12 y to the ends of its latus rectum

Summary:

The area of the triangle is 18 unit² while the vertex of the parabola x² = 12 y to the ends of its latus rectum