P.T the area of the triangle inscribed in the parabola y² = 4ax is $\frac{1}{8\mathrm{a}}\left|\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)\right|$ sqare unit 
where y₁, y₂, y₃ are the ordinates of its vertices.

Given parabola y² = 4ax
Let A(x₁, y₁),B(x₂ , y₂ ),C(x₃, y₃ ) be there points on the parabola y² = 4ax
$$\begin{gathered} {\mathrm{y}}_1^2=4{\mathrm{ax}}_1 \\ \Rightarrow {\mathrm{x}}_1=\frac{{\mathrm{y}}_1^2}{4\mathrm{a}},{\mathrm{x}}_2=\frac{{\mathrm{y}}_2^2}{4\mathrm{a}} \mathrm{and} {\mathrm{x}}_3=\frac{{\mathrm{y}}_3^2}{4\mathrm{a}} \end{gathered}$$
$\therefore \mathrm{A}\left(\frac{{\mathrm{y}}_1^2}{4\mathrm{a}},{\mathrm{y}}_1\right),\mathrm{B}\left(\frac{{\mathrm{y}}_2^2}{4\mathrm{a}},{\mathrm{y}}_2\right),\mathrm{C}\left(\frac{{\mathrm{y}}_3^2}{4\mathrm{a}},{\mathrm{y}}_3\right)$ are the points on parabola y² = 4ax
Area of Triangle $=Absolute value of\frac{1}{2}\left|\begin{array}{l}{\mathrm{x}}_1-{\mathrm{x}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}_1-{\mathrm{y}}_2\\ {\mathrm{x}}_2-{\mathrm{x}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{y}}_2-{\mathrm{y}}_3\end{array}\right|$
$\begin{array}{l}=Absolute value of \frac{1}{2}\left|\begin{array}{ll}\frac{{\mathrm{y}}_1^2-{\mathrm{y}}_2^2}{4\mathrm{a}}& {\mathrm{y}}_1-{\mathrm{y}}_2\\ \frac{{\mathrm{y}}_2^2-{\mathrm{y}}_3^2}{4\mathrm{a}}& {\mathrm{y}}_2-{\mathrm{y}}_3\end{array}\right|\\ =Absolute value of\frac{1}{2}\frac{1}{4\mathrm{a}}\left|\begin{array}{ll}{\mathrm{y}}_1^2-{\mathrm{y}}_2^2& {\mathrm{y}}_1-{\mathrm{y}}_2\\ {\mathrm{y}}_2^2-{\mathrm{y}}_3^2& {\mathrm{y}}_2-{\mathrm{y}}_3\end{array}\right|\\ =Absolute value of\frac{1}{8\mathrm{a}}\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)\left|\begin{array}{ll}{\mathrm{y}}_1+{\mathrm{y}}_2& 1\\ {\mathrm{y}}_2+{\mathrm{y}}_3& 1\end{array}\right|\\ =Absolute value of\frac{1}{8\mathrm{a}}\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)\left|{\mathrm{y}}_1-{\mathrm{y}}_3\right|\\ =\frac{1}{8\mathrm{a}}\left|\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)\right|\text{ sq. units }\end{array}$