Question

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 is-

• A. $16$ sq. units
• B. $12$ sq. units
• C. $18$ sq. units
• D. $24$ sq. units

Answer 1

Answer: (C)

$18$ sq. units

Given,

$x^2=12y$

$\Rightarrow 4a=12$

$\therefore a=3$

at $y=3,x^2=12\left(3\right)=36$

$\therefore x=\pm 6$

Therefore the coordinates are,

$A(-6,3),O(0,0),B(6,3)$

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

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

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

$=18sq.units$