Question

Let $A\left(a{t}_1^2,2a{t}_1\right);B\left(a{t}_2^2,2a{t}_2\right);C\left(a{t}_3^2,2a{t}_3\right)$ be three points on the parabola $y^2=4ax$. If the orthocentre of the triangle $ABC$ is at the focus, then find $\left|t_1+t_2+t_3+t_1{t}_2{t}_3+t_1{t}_2+t_2{t}_3+t_3{t}_1\right|$.

Answer 1

Answer: 5

$\frac{2}{t_1+t_2}\times \frac{2a{t}_3}{a\left(t_3^2-1\right)}=-1$
$\frac{2}{t_1+t_2}=\frac{1-t_3^2}{2t_3}$
$t_1+t_2=\frac{4t_3}{1-t_3^2}$
$t_1+t_2+t_3=\frac{4t_3}{1-t_3^2}+t_3=\frac{4t_3+t_3-t_3^3}{1-t_3^2}=\frac{5t_3-t_3^3}{1-t_3^2}$
If $t_1+t_2+t_3=u$ then $u=\frac{5t_3-t_3^3}{1-t_3^2}$ .....(1)
The Equation (1) suggests that $t_3$ is a root of
$t^3-u{t}^2-5t+u=0$ ....(2)
by symmetry $t_1$ and $t_2$ are also the root of (2). Hence
$t_1+t_2+t_3=u$
and $t_1{t}_2{t}_3=-u$
Hence $t_1+t_2+t_3+t_1{t}_2{t}_3=0$
also $t_1{t}_2+t_2{t}_3+t_3{t}_1=-5$
Hence $\left|t_1+t_2+t_3+t_1{t}_2{t}_3+t_1{t}_2+t_2{t}_3+t_3{t}_1\right|=5$