Parabola standard equation

Definitions. For a parabola, the fixed point is the focus, the fixed line is the directrix, the line through the focus perpendicular to the directrix is the axis, and the point where the curve crosses its axis is the vertex. Every point $P$ on the curve satisfies one rule: its distance from the focus equals its distance from the directrix.

Setting up coordinates. Place the vertex at the origin and let the axis lie along the positive $x$-axis. If the focus is $S(a,0)$ with $a>0$, the directrix must sit the same distance $a$ on the opposite side of the vertex, so it is the line $x=-a$. This placement is what makes the algebra collapse into a short polynomial rather than a messier expression.

The distance condition. For $P(x,y)$ on the parabola, the defining equality is

where: $S(a,0)$ is the focus, $x=-a$ is the directrix, and $x+a$ is the perpendicular distance from $P(x,y)$ to that vertical line (valid since every point of the parabola has $x\ge 0$, so $x+a>0$).

Four orientations, one shape parameter. Squaring and simplifying this condition (done in full in the derivation below) gives $y^2=4ax$. Reflecting the setup across the $y$-axis or the $x$-axis gives three more standard forms:

In every row the vertex stays at the origin and $a$ is both the vertex-to-focus distance and the vertex-to-directrix distance — the single number that fixes the whole shape. A larger $a$ pushes the focus further out and flattens the curve near the vertex; a smaller $a$ pulls the focus in and narrows the curve.

Symmetry as a sanity check. In $y^2=4ax$, swapping $y\to -y$ leaves the equation unchanged, so the curve is symmetric about the $x$-axis — consistent with the axis being the $x$-axis. Likewise $x^2=4ay$ is unchanged under $x\to -x$, giving symmetry about the $y$-axis. Whichever variable is squared tells you the axis; whichever variable appears alone (to the first power) tells you which direction the curve opens.

Why the vertex sits exactly midway. For any of the four forms, the vertex lies precisely halfway between the focus and the directrix along the axis. This is not a coincidence of the algebra — it follows directly from the defining condition, since the vertex is itself a point on the curve, so its distance to the focus must equal its distance to the directrix, and the only point on the axis with equal distances to $S(a,0)$ and to $x=-a$ is the origin. Every other point on the parabola trades some of that balance: as $P$ moves away from the vertex, both distances grow together, but always by exactly matching amounts, which is what keeps the curve open and unbounded rather than closing back on itself like an ellipse.

Problem. Find the equation of the parabola with vertex at the origin, axis along the $x$-axis, opening rightward, and passing through the point $(2,-4)$. Then verify the point satisfies the focus-directrix condition directly.

Step 1. Since the parabola opens right with vertex at the origin, its equation has the form $y^2=4ax$.

Step 2. Substitute the point $(2,-4)$: $(-4)^2 = 4a(2) \Rightarrow 16 = 8a \Rightarrow a=2$.

Step 3. So the equation is $y^2 = 4(2)x = 8x$, with focus $(2,0)$ and directrix $x=-2$.

Step 4 (verification). Distance from $(2,-4)$ to the focus $(2,0)$ is $\sqrt{(2-2)^2+(-4-0)^2}=4$. Distance from $(2,-4)$ to the directrix $x=-2$ is $2-(-2)=4$. Both distances equal $4$, confirming the point genuinely lies on $y^2=8x$.

Answer: $y^2=8x$, focus $(2,0)$, directrix $x=-2$.

Competitive stretch

For the parabola $x^2=-12y$: matching to $x^2=-4ay$ gives $4a=12\Rightarrow a=3$, so vertex $(0,0)$, focus $(0,-3)$, directrix $y=3$. Since the coefficient of $y$ is negative the curve opens downward, so the focus should sit below the vertex and the directrix above it — exactly what the values show, a quick self-check worth running on every orientation problem.

• Mixing up which variable is squared for which orientation — $y^2=4ax$ opens along the $x$-axis, while $x^2=4ay$ opens along the $y$-axis; squaring the wrong variable flips the whole picture.
• Forgetting the negative sign for leftward or downward-opening parabolas, e.g. writing $y^2=4ax$ when the parabola actually opens left (it should be $y^2=-4ax$).
• Confusing $a$ in $4a$ with the coefficient itself — for $y^2=12x$, students often say $a=12$ instead of correctly solving $4a=12\Rightarrow a=3$.
• Assuming the vertex is always the origin — that is only true for these standard forms; a shifted parabola is outside this syllabus's coordinate-derivation scope.

Ask the tutor: Ask the AI tutor to re-derive $y^2=4ax$ with you step by step from the distance condition, or to walk through why the four orientations are just reflections of one master derivation.