Question

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.

Answer 1

The given parabola is $x^2=12y$

On comparing this equation with $x^2=4ay$, we get

$\Rightarrow 4a=12$

$\Rightarrow a=3$

$\Rightarrow$So, the coordinates of focus is $S(0,a)=S(0,3)$

$\Rightarrow$Let $AB$ be the latus rectum of the given parabola.

$\Rightarrow$Coordinates of end-points of latus rectum are $(-2a,a),(2a,a)$

$\Rightarrow$Coordinates of $A$ are $(-6,3)$ while the coordinates of $B$ are $(6,3)$

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

$\Rightarrow$Let $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ be the coordinates of $\Delta OAB$

$\Rightarrow$Area of $△$ $OAB$

$=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)|$

$=\frac{1}{2}|0(3-3)+(-6)(3-0)+6(0-3)|$

$=\frac{1}{2}|(-6)\left(3\right)+6(-3)|$

$=\frac{1}{2}|-18-18|$

$=\frac{1}{2}|-36|$

$=\frac{1}{2}\times 36$

$=18$ sq.units

Thus the required area of the triangle is $18$ unit${}^2$