Equation of the Normal and the Number of Normals

The normal to the parabola at a point is the line through that point perpendicular to the tangent there, and once the tangent's slope is known, the normal's equation is immediate. At $(x_1,y_1)$, the tangent $S_1=0$, i.e. $yy_1=2a(x+x_1)$, has slope $\dfrac{2a}{y_1}$; the perpendicular line therefore has slope $-\dfrac{y_1}{2a}$, giving

Substituting the parametric point $(at^2,2at)$ for $(x_1,y_1)$ and simplifying turns this into the compact parametric form

It is also worth having the normal in slope form. If the normal at parameter $t$ has slope $m$, then from $y+xt=2at+at^3$ the slope is $-t$, so $m=-t \Rightarrow t=-m$; substituting back gives the normal of slope $m$ as

which touches (in the sense of passing through) the curve at the point $t=-m$, i.e. at $(am^2,\,-2am)$. This slope form is exactly analogous to the tangent's $y=mx+a/m$, and the two are used in mirror-image ways: the tangent formula answers "what line of slope $m$ touches the curve", while the normal formula answers "what line of slope $m$ meets the curve perpendicularly at one of its points".

A natural question is: given an external point $(x_1,y_1)$, how many normals to the parabola pass through it? Substituting $(x_1,y_1)$ into the parametric normal equation gives $y_1+x_1t = 2at+at^3$, i.e.

a cubic in $t$. A cubic with real coefficients always has at least one real root and at most three, so the number of normals from a given point to a parabola is 1, 2, or 3, depending on how many real roots this cubic has (found in practice using the discriminant of a depressed cubic, or, in simple cases, by inspection). This is a genuinely different count from the tangent case: exactly two tangents always exist from any external point, but the normal count depends on exactly where the point sits relative to the curve.

Worked example. Find the equation of the normal to $y^2=12x$ at the point where $t=1$, and verify it using the Cartesian point form. …