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The area of the triangle inscribed in the parabola\mathrm{y^2=4 a x} is 

\mathrm{\left|\mathrm{k}\left(t_1-t_2\right)\left(t_2-t_3\right)\left(t_3-t_1\right)\right| \text { where } t_1, t_2 \text { and } t_3}  are the parameters of the vertices, where k =

 

Option: 1

a


Option: 2

\mathrm{a^{2}}


Option: 3

\mathrm{a/2}


Option: 4

\mathrm{2a^{2}}


Answers (1)

best_answer

The three points on the parabola are \mathrm{\left(a t_1^2, 2 a t_1\right),\left(a t_2^2, 2 a t_2\right) \text { and }\left(a t_3^2, 2 a t_3\right)}

\mathrm{\text { Hence, area } \Delta=\left|\frac{1}{2}\right| \begin{array}{lll} a t_1^2 & 2 a t_1 & 1 \\ a t_2^2 & 2 a t_2 & 1 \\ a t_3^2 & 2 a t_3 & 1 \end{array}||=\left|a^2\right| \begin{array}{ccc} t_1^2 & t_1 & 1 \\ t_2^2-t_1^2 & t_2-t_1 & 0 \\ t_3^2-t_1^2 & t_3-t_1 & 0 \end{array}||}

\mathrm{=\left|a^2\left(t_1-t_2\right)\left(t_2-t_3\right)\left(t_3-t_1\right)\right| .}

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