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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

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

$\begin {array} {1 1} (A)\;16\: sq.units & \quad (B)\;17\: sq.units \\ (C)\;18\: sq.units & \quad (D)\;19\: sq.units \end {array}$

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A)
Toolbox:
  • Standard equation of a parabola which is open rightward is $ y^2=4ax$ where a is the focus.
Step 1 :
The given equation of the parabola is
$x^2=12y$
Now comparing this with the general equation $x^2=4ay$ we get,
$4a=12 \Rightarrow a = 3$
$ \therefore $ The coordinates of foci are $F(0, a)$. Let AB be the latus rectum of the given parabola.
Step 2
At $y=3, x^2=12(3)$
$ \Rightarrow x^2=36$
$ \therefore x = \pm 6$
$ \therefore $ The coordinates of A are (-6, 3) and coordinates of B are (6,3)
$ \therefore $ The vertices of $ \Delta 0AB $ are
$ 0(0,0), A ( -6, 3) \: and \: B(6,3)$
Step 3 :
Area of the triangle is $ \large\frac {1}{2}$$ \bigg| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \bigg|$
Area of a triangle ABC, whose coordinates are $(x_1, y_1), (x_2, y_2), (x_3, y_3)$
Substituting the values we get,
area of $ \Delta 0AB$
$ \large\frac{1}{2} $$ \bigg| 0(3-3)+(-6)(3-0)+6(0-3) \bigg| $ sq.units
$ = \large\frac{1}{2}$$ \bigg| -36-36 \bigg|$ sq.units
$ = \large\frac{1}{2}$$ \bigg| -36 \bigg|$ sq.units
$ = \large\frac{1}{2}$$ \times 36$ sq.units
= 18 sq.units
Hence the required area of the triangle is 18 sq.units.
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