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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.

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