Normal with Given Slope

Normal to y² = 4ax with slope m. Up to three normals can be drawn from a given external point, corresponding to three real roots of the cubic in m.

The normal at parameter $t$ is $y + tx = 2at + at^3$, with slope $-t$. Let the slope of the normal be $m = -t$, so $t = -m$.

Substituting $t = -m$:

$$y + (-m)x = 2a(-m) + a(-m)^3$$ $$y - mx = -2am - am^3$$ $$y = mx - 2am - am^3$$

This is the normal with slope $m$ to $y^2 = 4ax$.

Point of contact: The point with parameter $t = -m$ is $(am^2, -2am)$.

Three normals from a point $(h, k)$: Substituting $(h, k)$ into the normal equation:

$$k = mh - 2am - am^3$$ $$am^3 + m(2a - h) + k = 0 \quad \cdots (*)$$

This cubic in $m$ gives the slopes of all normals from $(h, k)$.

By Vieta's formulas for the three roots $m_1, m_2, m_3$:

$$m_1 + m_2 + m_3 = 0$$ $$m_1m_2 + m_2m_3 + m_3m_1 = \frac{2a - h}{a}$$ $$m_1m_2m_3 = -\frac{k}{a}$$

The cubic (*) has three real roots (three real normals from $(h,k)$) when a certain discriminant condition holds. Otherwise, one real normal and two complex ones.