# Area of triangle with two of its vertex at the latus rectu.

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

Attempt-

$$x^{2}=12x$$ $$a = 3$$ $$\text{Focus } = (0,3)$$ Let, latus rectum be PQ passing through focus S.

Now, I tried applying distance formula for PS=SQ but it doesn't get any result and also I tried the distance formula on PQ and equate it to 2k. Neither worked, if only I could get (h,k) one end of latus rectum, I could solve by Herons formula but I am not able to find the coordinates (h,k). Please advise.

Since $a=3$, y coordinate of $P$ and $Q$ is 3, putting that into the equation we get $P=(6,3)$ and $Q=(-6,3)$. Now area of required triangle is $1/2 \times PQ \times OS$, where O is vertex. Note $OS=3$ and $PQ=12$.