General (Shifted) Form of a Parabola

Shifting the vertex. All the standard forms so far have their vertex fixed at the origin. If instead the vertex is shifted to a general point $(h,k)$, while the axis of symmetry stays parallel to a coordinate axis, the equation becomes (for the case of an axis parallel to the $X$-axis):

This represents a parabola whose:

• axis of symmetry is the line $y-k=0$ (parallel to the $X$-axis),
• vertex is at $(h,k)$,
• focus is at $(h+a,\,k)$,
• directrix is $x=h-a$.

Expanding the square, $(y-k)^2=4a(x-h)$ can always be rearranged into the form $x = Ay^2+By+C$ for suitable constants $A,B,C$ — this is a useful way to recognise a shifted parabola when it is given to you in expanded form: if an equation is quadratic in one variable and linear in the other, it is a parabola, and completing the square recovers $h,k,a$.

Equivalently, writing $X=x-h,\ Y=y-k$ (a simple change of variables that moves the origin to the new vertex), the equation becomes exactly the familiar standard form $Y^2=4aX$ — so every one of sections 7.1.6–7.1.9's results applies to the shifted parabola too, just translated by $(h,k)$.

Worked Example 1 — focus, directrix, latus rectum for two given parabolas.

1. $y^2=28x$. Comparing with $y^2=4ax$: $4a=28\Rightarrow a=7$. Focus $S(a,0)=(7,0)$; directrix $x+7=0$; latus rectum $=4a=28$; end points $(7,14)$ and $(7,-14)$.
2. $3x^2=8y$, i.e. $x^2=\dfrac83y$. Comparing with $x^2=4by$: $4b=\dfrac83\Rightarrow b=\dfrac23$. Focus $\left(0,\dfrac23\right)$; directrix $3y+2=0$; latus rectum $=4b=\dfrac83$; end points $\left(\dfrac43,\dfrac23\right)$ and $\left(-\dfrac43,\dfrac23\right)$. Worked Example 2 — equation from vertex, axis and a point. Vertex at the origin, axis along $Y$, through $(6,-3)$: form is $x^2=4by$. Substituting: $36=4b(-3)\Rightarrow -12b=36\Rightarrow b=-3$. Equation: $x^2=-12y$, i.e. $x^2+12y=0$. Worked Example 3 — equation from the directrix alone. Directrix $x+3=0$, so comparing with $x+a=0$ gives $a=3$. Since the vertex is (by default, unless stated otherwise) the origin and the parabola opens toward the focus (away from the directrix): equation is $y^2=4ax=12x$. Worked Example 4 — focal distance from the ordinate. Parabola $y^2=20x$ ($4a=20\Rightarrow a=5$); a point has ordinate ($y$-coordinate) $10$. Then $100=20x\Rightarrow x=5$. Focal distance $=a+x=5+5=10$ units. Worked Example 5 — equation from one extremity of the latus rectum. Given $(4,-8)$ as one extremity of the latus rectum: the other extremity must be its mirror image $(4,8)$ (latus-rectum end points are symmetric about the axis). Matching $(a,\pm2a)=(4,\pm8)$ gives $a=4$. Equation: $y^2=4(4)x=16x$. …