Parabola — Equation, Properties & Latus Rectum

With vertex at the origin and $a > 0$:

• $y^2 = 4ax$: opens right, focus $(a,0)$, directrix $x=-a$, axis $y=0$.
• $y^2 = -4ax$: opens left, focus $(-a,0)$, directrix $x=a$.
• $x^2 = 4ay$: opens up, focus $(0,a)$, directrix $y=-a$.
• $x^2 = -4ay$: opens down, focus $(0,-a)$, directrix $y=a$.

The squared variable names the axis; the sign of the linear term gives the direction. From a focus and directrix, the vertex is their midpoint. A parabola described by its slope: "the tangent at every point has slope $\dfrac{2y}{x}$" is the separable differential equation $\dfrac{dy}{dx}=\dfrac{2y}{x}\Rightarrow\dfrac{dy}{y}=\dfrac{2\,dx}{x}\Rightarrow\ln y=2\ln x+\ln c\Rightarrow y=cx^2$ — the family of parabolas $x^2=4ay$ with vertex at the origin. Likewise $\dfrac{dy}{dx}=\dfrac{y}{2x}$ gives $y^2=cx$. (Separation of variables is in the Differential Equations notes.)

For $y^2 = 4ax$, the latus rectum is the vertical chord through the focus $(a,0)$, with length $4a$ and endpoints $(a, \pm 2a)$. Knowing the latus rectum's endpoints fixes the focus (their midpoint) and the value of $a$; two parabolas can share the same latus rectum (opening in opposite directions).

For a point $(x_1, y_1)$ on $y^2 = 4ax$:

• Focal distance $= x_1 + a$ (the distance to the directrix $x=-a$, by the focus–directrix property).
• For $x^2 = 4ay$, the focal distance is $y_1 + a$.
• A focal chord passes through the focus; the latus rectum is the shortest focal chord.

• Tangent of slope $m$ to $y^2 = 4ax$: $y = mx + \dfrac{a}{m}$ (touch point $\left(\tfrac{a}{m^2}, \tfrac{2a}{m}\right)$). A tangent inclined at angle $\theta$ has $m=\tan\theta$.
• Chord through the vertex at angle $\theta$: substitute $y = x\tan\theta$ to find where it meets the curve.
• Two parabolas $y^2=4ax$ and $x^2=4ay$ meet at $(0,0)$ and $(4a,4a)$, both on the line $y=x$.
• Length of a chord through the vertex: the chord at angle $\theta$ meets $y^2=4ax$ again at $(4a\cot^2\theta,\,4a\cot\theta)$, so its length is $4a\cot\theta\csc\theta$ (distance formula plus $1+\cot^2\theta=\csc^2\theta$); the focal chord at angle $\theta$ has length $4a\csc^2\theta$.
• Equilateral triangle inscribed with one vertex at the vertex of $y^2=4ax$: by symmetry the other two vertices are mirror images across the axis, so the sides through the vertex make $30°$ with the axis: $y=\pm x\tan30°=\pm\tfrac{x}{\sqrt3}$. Substituting, $x=12a$, $y=\pm4\sqrt3a$, so the side is $8\sqrt3\,a$.