Question

If $y_1,y_2$ and $y_3$ are the ordinates of the vertices of a triangle inscribed in the parabola $y^2=4ax$, then its area is

• A. $\frac{1}{2a}(y_1-y_2)(y_2-y_3)(y_3-y_1)$
• B. $\frac{1}{4a}(y_1-y_2)(y_2-y_3)(y_3-y_1)$
• C. $\frac{1}{8a}(y_1-y_2)(y_2-y_3)(y_3-y_1)$
• D. none of these

Answer 1

The correct option is C $\frac{1}{8a}(y_1-y_2)(y_2-y_3)(y_3-y_1)$

Let $x_1,x_2$ and $x_3$ be the abscissae of the points on the parabola whose ordinates are $y_1,y_2$ and $y_3$, respectively. Then $y_1^2=4a{x}_1,y_2^2=4a{x}_2$ and $y_3=4a{x}_3$. Therefore, the area of the triangle whose vertices are $(x_1,y_1)(x_2,y_2)$ and $(x_3,y_3)$ is

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}\frac{y_1^2}{4a}& y_1& 1\\ \frac{y_2^2}{4a}& y_2& 1\\ \frac{y_3^2}{4a}& y_3& 1\end{array}\right|$

$=\frac{1}{8a}\left|\begin{array}{ccc}{y}_1^2& y_1& 1\\ y_2^2& y_2& 1\\ y_3^2& y_3& 1\end{array}\right|$

$=y_1^2(y_2-y_1)-y_1(y_2^2-y_3^2)+(y_2^2{y}_3-y_2{y}_3^2)$

$=\frac{1}{8a}(y_1-y_2)(y_2-y_3)(y_3-y_1)$