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

Asked by Sakshi | 9 months ago |  88

##### Solution :-

The given parabola is x2 = 12y.

On comparing this equation with x2 = 4ay, we get,

4a = 12

a =$$\dfrac{12}{4}$$

= 3

The coordinates of foci are S(0,a) = S(0,3).

Now 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

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

Then, the vertices of ΔOAB are O(0,0), A (-6,3) and B(6,3).

By using the formula,

Area of ΔOAB = $$\dfrac{1}{2}$$ [0(3-3) + (-6)(3-0) + 6(0-3)] unit2

= $$\dfrac{1}{2}$$ [(-6) (3) + 6 (-3)] unit2

= $$\dfrac{1}{2}$$ [-18-18] unit2

= $$\dfrac{1}{2}$$ [-36] unit2

= 18 unit2

Area of ΔOAB is 18 unit2

Answered by Aaryan | 9 months ago

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