# Prove that normals at the points where line intersects a parabola meet at point on normal at a given point on the parabola.(Loney XXX 18)

Prove that the normals at the points, where the straight line $lx+my=1$ meets the parabola, meet on the normal at the point $(\frac{4am^2}{l^2},\frac{4am}{l})$ of the parabola.
As the line intersects the parabola we have $$y^2=4a \Bigg( {\frac{1-my}{l}}\Bigg)$$ hence $$y=\frac{-4amy \pm 4 \sqrt{a^2m^2-a^2l^2}}{2l}$$ Now though I know how to find equation of a normal from a given point the expressions henceforth become creepy. There must be a other way round to solve this problem. How do I get that ?

Let $\left(\frac{s^2}{4a},s\right),\left(\frac{t^2}{4a},t\right)$ where $s\not= t$ be the intersection points between the parabola $y^2=4ax$ and the line $lx+my=1$.

Now, the equation of the normal at the point $\left(\frac{s^2}{4a},s\right),\left(\frac{t^2}{4a},t\right)$ of the parabola is $$y-s=-\frac{s}{2a}\left(x-\frac{s^2}{4a}\right)\tag1$$ $$y-t=-\frac{t}{2a}\left(x-\frac{t^2}{4a}\right)\tag2$$ respectively. So, the intersection point between $(1)$ and $(2)$ is $$\left(2a+\frac{(s+t)^2-st}{4a},-\frac{st(s+t)}{8a^2}\right)\tag3$$

By the way, eliminating $x$ from $y^2=4ax$ and $lx+my=1$ gives $$y^2+\frac{4am}{l}y-\frac{4a}{l}=0.$$ By Vieta's formulas, we have $$s+t=-\frac{4am}{l},\quad st=-\frac{4a}{l}.$$ (Note here that we may suppose that $l\not=0$.) Hence, we know that $(3)$ can be written as $$\left(2a+\frac{(-\frac{4am}{l})^2-(-\frac{4a}{l})}{4a},-\frac{-\frac{4a}{l}(-\frac{4am}{l})}{8a^2}\right),$$ i.e. $$\left(2a+\frac{4am^2+l}{l^2},-\frac{2m}{l^2}\right)\tag4$$

Now, the equation of the normal at the point $\left(\frac{4am^2}{l^2},\frac{4am}{l}\right)$ of the parabola is $$y-\frac{4am}{l}=-\frac{1}{2a}\cdot\frac{4am}{l}\left(x-\frac{4am^2}{l^2}\right)\tag5$$ Here, we can see that $(4)$ is on $(5)$.

if the Normals of the parabola $y^2=4ax$ at the points $t_1$,$t_2$ and $t_3$ are Concurrent then we can easily prove that $$t_1+t_2+t_3=0$$

Now assuming $t_1$ and $t_2$ Lie on $lx+my=1$ we have

$$lat_1^2+2amt_1=1$$ and

$$lat_2^2+2amt_2=1$$ Subtracting above equations we get

$$la(t_1+t_2)+2am=0$$ $\implies$

$$-lat_3=-2am$$ So

$$t_3=\frac{2m}{l}$$

Hence the Normals at two points $t_1$ and $t_2$ meet the normal at $t_3$ whose coordinates are

$$\left(at_3^2,2at_3\right)=\left(\frac{4a^2m^2}{l^2},\frac{4am}{l}\right)$$