3

The normal at $A$ and $B$ on the parabola $y^2=4ax$ meet the parabola at $C$ on same parabola.

Then locus of orthocenter of $\triangle ABC$

Attempt Let $A(at^2_{1},2at_{1})$ and $B(at^2_{2},2at_{2})$. Then equation of Normal $A$ and $B$ is

$y=-t_{1}x+2at_{1}+at^3_{1}$ and $y=-t_{2}x+2at_{2}+at_{2}^3$

So coordinate of point $C\bigg(a(2+t^2_{1}+t^2_{2}+t_{1}t_{2}),at_{2}(t_{1}+t_{2})\bigg)$

could some help me how to solve it, thanks

Community
  • 1
DXT
  • 11,241
  • Your point $C$ ordinate is wrong – Your IDE Dec 15 '17 at 05:20
  • You appear to be asserting that $C$ is on the parabola for arbitrary values of $t_1$ and $t_2$. This is not the case. – amd Dec 15 '17 at 06:44
  • 2
    Another point of departure would be to use the evolute (envelope of the normals). See (http://mathworld.wolfram.com/ParabolaEvolute.html). – Jean Marie Dec 28 '17 at 13:59

1 Answers1

2

You must start with point $C=(y_C^2/4a,y_C)$: the normal at a generic point $(y^2/4a,y)$ on the parabola passes through $C$ if: $$ y-y_C=-{y\over2a}{y^2-y_C^2\over4a}, \quad\hbox{that is:}\quad y={-y_C\pm\sqrt{y_C^2-32a^2}\over2}. $$ These two solutions above give then the coordinates of $A$ and $B$ (provided of course $y_C^2\ge32a^2$): $$ y_A={-y_C+\sqrt{y_C^2-32a^2}\over2},\quad y_B={-y_C-\sqrt{y_C^2-32a^2}\over2},\quad $$ and of course $x_A=y_A^2/4a$, $x_B=y_B^2/4a$.

Now that you know the coordinates of $A$, $B$ and $C$ (as a function of $y_C$) I think you can go on by yourself to find the orthocenter of triangle $ABC$, whose locus should lie on another parabola.

EDIT.

If $O=(x_O,y_O)$ is the orthocenter, a straightforward computation gives in fact:

$$ x_O=x_C-6a,\quad y_O=-{y_C\over2}, $$ and $O$ lies then on the parabola of equation $y^2=ax+6a^2$.

Intelligenti pauca
  • 50,470
  • 4
  • 42
  • 77