Prove that the orthocentres of the triangles formed by three tangents and the corresponding three normals to a parabola are equidistant from the axis.

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Solution

Let the parabola be $y^{2} = 4 a x . . . . . (1)$

and $(a m_{1}^{2}, - 2 a m_{1}), (a m_{2}^{2}, - 2 a m_{2})$ and $(a m_{3}^{2}, - 2 a m_{3})$ be the coordinates of points P,Q,R respectively
Then, the equation of the tangents at P,Q andR with respect to the parabola (I) are

$- m_{1} y = x + a m_{1}^{2}$ ...... $(2)$

$- m_{2} y = x + a m_{2}^{2}$ ....... $(3)$

$- m_{3} y = x + a m_{3}^{2}$ ........ $(4)$

Solving $(2)$ and $(3)$ we have the coordinates of the points of the points of intersection say $A$ as

$(a m_{1} m_{2}, - a (m_{1} + m_{2}))$

Now equation to the line through $A$ and perpendicular to $(3)$ will be

$y + a (m_{1} + m_{2}) = m_{3} (a m_{1} m_{3})$ ........... $(5)$

Similarly the equation to the line through $B$ the point of intersection of $(3)$ and $(4)$ and perpendicular to $(2)$ is
$y + a (m_{2} + m_{3}) = m_{1} (x - a m_{2} m_{3})$ ......... $(6)$
On solving $(5)$ and $(6)$ we get the ordinate of the triangle formed by $(2), (3)$ and $(4)$ as
$- a (m_{1} + m_{2} + m_{3} + m_{1} m_{2} m_{3})$
Again equation to the normal at P,Q&R are respectively
$y = m_{1} x - 2 a m_{1} - a m_{1}^{3}$ .......................(7)
$y = m_{2} x - 2 a m_{2} - a m_{2}^{3}$ .......................(8)
$y = m_{3} x - 2 a m_{3} - a m_{3}^{3}$ .......................(9)
On solving $(8)$ and $(7)$ we get the coordinate of one of the vertices of the triangle formed by these three normals as
$2 a + a (m_{1}^{2} + m_{2}^{2} + m_{1} m_{2}), a m_{1} m_{2} (m_{1} + m_{2})$
Equation to the line through this point and perpendicular to (9) is
$y - a m_{1} m_{2} (m_{1} + m_{2}) = - \frac{1}{m_{2}} (x - 2 a - a m_{1}^{2} - a m_{2}^{2} - a m_{1} m_{2})$ ...........(10)
Similarly the equation to the other perpendicular will be
$y - a m_{2} m_{3} (m_{2} + m_{3}) = - \frac{1}{m_{1}} (x - 2 a - a m_{2}^{2} - a m_{3}^{2} - a m_{2} m_{3})$ ...........(11)
On solving $(10)$ and $(11)$ we get the ordinate of the point of intersection as $- a (m_{1} + m_{2} + m_{3} + m_{1} m_{2} m_{3})$
This is also the ordinate of the point of intersection or orthocentre of the triangle formed by tangents.
Hence the ortho centres are equidistant from the axis that is x-axis

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