The area of triangle formed by the lines joining the vertex of the parabola, $x^2=8 y$, to the extremities of its latus rectum is

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Aicharya · Accepted Answer

Given parabola is $\mathrm{x}^2=8 \mathrm{y}$ $\Rightarrow 4 a=8 \Rightarrow a=2$ To find: Area of $\triangle A B C$ $ \begin{aligned} & A=(-2 a, a)=(-4,2) \\ & B=(2 a, a)=(4,2) \\ & \mathrm{C}=(0,0) \end{aligned} $

$ \begin{aligned} & \therefore \text { Area }=\frac{1}{2}\left|\begin{array}{ccc} -4 & 2 & 1 \\ 4 & 2 & 1 \\ 0 & 0 & 1 \end{array}\right| \\ & =\frac{1}{2}[-4(2)-2(4)+1(0)] \\ & =\frac{-16}{2}=-8 \approx 8 \text { sq. unit } \\ & (\because \text { area cannot be negative }) \end{aligned} $

The area of triangle formed by the lines joining the vertex of the parabola, $x^2=8 y$, to the extremities…

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