Question
What is the area of the triangle formed by the lines joining the vertex of the parabola $${x^2} = 12y$$ to the ends of the latus rectum ?
A.
$$9$$ square units
B.
$$12$$ square units
C.
$$14$$ square units
D.
$$18$$ square units
Answer :
$$18$$ square units
Solution :
Given parabola $${x^2} = 12y$$ is which is of the form $${x^2} = 4ay$$
$$ \Rightarrow 4a = 12 \Rightarrow a = 3$$
Now, $$LM$$ is the latus rectum whose length $$ = 4a = 4 \times 3 = 12$$
So, area of $$\Delta LMV$$
$$\eqalign{ & = \frac{1}{2} \times LM \times VF \cr & = \left( {\frac{1}{2} \times 12 \times 3} \right){\text{square units}} \cr & = 18{\text{ square units}} \cr} $$
Given parabola $${x^2} = 12y$$ is which is of the form $${x^2} = 4ay$$
$$ \Rightarrow 4a = 12 \Rightarrow a = 3$$
Now, $$LM$$ is the latus rectum whose length $$ = 4a = 4 \times 3 = 12$$
So, area of $$\Delta LMV$$
$$\eqalign{ & = \frac{1}{2} \times LM \times VF \cr & = \left( {\frac{1}{2} \times 12 \times 3} \right){\text{square units}} \cr & = 18{\text{ square units}} \cr} $$