Condition for a Tangent and the Equation of the Tangent

A line and a parabola can meet in two points, touch at exactly one point (a tangent), or miss the curve entirely, and all three cases fall straight out of solving them simultaneously. Substitute the line $y=mx+c$ (with slope $m\ne 0$) into $y^2=4ax$:

This is a quadratic in $x$, so its roots (the $x$-coordinates of the intersection points) are two distinct real numbers, one repeated real number, or a complex-conjugate pair, according to whether the discriminant is positive, zero, or negative. The line is a tangent exactly when the two intersection points coincide, i.e. when the discriminant is zero:

So $y=mx+c$ touches $y^2=4ax$ exactly when $c=a/m$ — meaning that for every nonzero slope $m$, the specific line

is automatically a tangent to the parabola, and its point of contact works out to $\left(\dfrac{a}{m^2},\ \dfrac{2a}{m}\right)$. Two side notes fall out immediately: a horizontal line ($m=0$) can never be tangent to $y^2=4ax$ (it always cuts the curve in one point at $\left(\tfrac{c^2}{4a},c\right)$, never touching it), while the $y$-axis itself ($x=0$) is the one tangent not of the form $y=mx+c$ — it touches the curve only at the vertex. Because the tangent condition gives a genuine quadratic in $m$ for any external point (see below), exactly two tangents can always be drawn to a parabola from a point lying outside it.

The formula above is convenient for tangents of a given slope, but the more commonly needed formula is the tangent at a given point $(x_1,y_1)$ already known to lie on the curve. This follows from a slightly different route: the chord joining two points $(x_1,y_1)$ and $(x_2,y_2)$ on $S=0$ turns out to have the strikingly simple equation $S_1+S_2=S_{12}$ (both sides reduce to the same straight line $4ax-(y_1+y_2)y+y_1y_2=0$ once you use $y_1^2=4ax_1$ and $y_2^2=4ax_2$). Letting the second point slide along the curve until it merges into the first — the geometric meaning of "the chord becomes the tangent" — replaces $S_2$ by $S_1$ and $S_{12}$ by $S_{11}=0$, leaving $2S_1=0$. Hence:

In parametric form, replacing $(x_1,y_1)$ by $(at^2,2at)$ in $S_1=0$ and simplifying gives the equally compact

Worked example 1. Find the equation of the tangent to $y^2=16x$ at the point $(4,8)$, and separately find the tangent of slope $2$.

Here $4a=16$, so $a=4$. First check $(4,8)$ lies on the curve: $8^2=64=16(4)$, good. The tangent at a point uses $S_1=0$: $yy_1-2a(x+x_1)=0 \Rightarrow 8y - 8(x+4)=0 \Rightarrow 8y-8x-32=0 \Rightarrow y = x+4$. For the tangent of slope $m=2$: using $y=mx+a/m$ gives $y=2x+\tfrac{4}{2}=2x+2$, touching the parabola at $\left(\tfrac{a}{m^2},\tfrac{2a}{m}\right)=\left(1,4\right)$ — and indeed $4^2=16=16(1)$ confirms that point is on the curve.

Worked example 2. Show that the line $x+y+2=0$ is a tangent to $y^2=8x$ and find the point of contact. …