Question

Prove that the area of the triangle formed by the tangents at $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ to the parabola $y^2=4ax$ $(a>0)$ is $\frac{1}{16a}\left|(y_1-y_2)(y_2-y_3)(y_3-y_1)\right|$ sq.units.

Answer 1

$y^2=4ax$

$2y{y}^{\prime }=4a\Rightarrow y^{\prime }=\frac{2a}{y}$

So for $(x_n,y_n)$ tangent is

$y-y_n=\frac{2a}{y_n}(x-x_n)$

$y{y}_n-{y_n}^2=2ax-2a{x}_n$

$y{y}_n-4a{x}_n=2ax-2a{x}_n$

$y{y}_n=2ax+2a{x}_n$

$y{y}_n=2ax\frac{{y_n}^2}{2}$

So, $y{y}_1=2ax\frac{{y_1}^2}{2},y{y}_2=2ax\frac{{y_2}^2}{2},y{y}_3=2ax\frac{{y_3}^2}{2}$

Intersection of $\left(1\right)$ and $\left(2\right)$ is $y(y_1-y_2)=\frac{{y_1}^2}{2}-\frac{{y_2}^2}{2}$

$y=\frac{y_1+y_2}{2}$

So, $2ax=\frac{y_1+y_2}{2}\times y_1-\frac{{y_1}^2}{2}$

$=\frac{y_1{y}_2}{2}$

$x=\frac{y_1{y}_2}{4a}$

So, intersection of $\left(1\right)$ and $\left(2\right)$ is $(\frac{y_1{y}_2}{4a},\frac{y_1{+y}_2}{2})$

So, Area$=\left|\begin{array}{cc}{Y}_1-Y_2& X_1-X_2\\ Y_2-Y_3& X_2-X_3\end{array}\right|$

$=\left|\begin{array}{cc}\frac{y_2+y_3}{2}-\frac{y_1+y_3}{2}& \frac{y_2{y}_3}{4a}-\frac{y_1{y}_3}{4a}\\ \frac{y_1+y_3}{2}-\frac{y_1+y_2}{2}& \frac{y_1{y}_3}{4a}-\frac{y_1{y}_2}{4a}\end{array}\right|$

$=\left|\begin{array}{cc}\frac{y_2-y_1}{2}& \frac{y_3(y_2-y_1)}{4a}\\ \frac{y_3-y_2}{2}& \frac{y_1(y_3-y_2)}{4a}\end{array}\right|$

$=|\frac{y_1}{4a}(y_2-y_1)(y_3-y_2)-\frac{y_3}{4a}(y_2-y_1)(y_3-y_2)|$

$=\frac{1}{4a}\left|(y_1-y_3)(y_2-y_1)(y_3-y_2)\right|$