A parabola is the set of all points in a plane that are equidistant from a fixed point and a fixed straight line. The fixed point is the focus and the fixed line is the directrix. Writing $P$ for a point on the curve, $F$ for the focus and $M$ for the foot of the perpendicular from $P$ to the directrix, the defining condition is simply:
$$PF = PM$$
This balance between a point and a line is what gives the parabola its single, sweeping branch. (In the language of eccentricity, a parabola is the conic with $e = 1$, since $PF = e\cdot PM$ with $e = 1$.)
Key vocabulary. The axis is the line through the focus perpendicular to the directrix — the curve's line of symmetry. The vertex is the point where the parabola meets its axis, sitting exactly midway between focus and directrix. The focal distance of any point on the parabola is its distance from the focus. The latus rectum is the chord through the focus perpendicular to the axis; for all four standard forms below its length is the same:
$$\text{Length of latus rectum} = |4a|$$
The four standard parabolas. Placing the vertex at the origin and the axis along a coordinate axis gives four forms. The variable that is squared and the sign together tell you instantly which way the curve opens:
In every case $a > 0$ is the distance from the vertex to the focus (and equally from the vertex to the directrix). A clean rule of thumb: if $y$ is squared the parabola opens horizontally, if $x$ is squared it opens vertically; the sign then chooses the direction (positive $=$ right/up, negative $=$ left/down). The endpoints of the latus rectum of $y^2 = 4ax$ are $(a, \pm 2a)$ — a fact worth remembering for sketching.
Finding the equation from conditions. Decide the orientation from the data (where the focus or directrix sits, or which way the curve opens), pick the matching standard form, then find $a$. If you are given a point the curve passes through, substitute it to solve for $a$; if you are given the focus or directrix, read $a$ directly. When the focus and directrix straddle the origin symmetrically, the vertex is their midpoint — the origin — and a standard form applies.
Deeper Insight — why "equal distances" forces a curve, and where it shows up: The defining condition $PF = PM$ is deceptively simple, yet squaring it produces exactly a second-degree equation with only one squared variable — which is what makes a parabola distinct from a circle or an ellipse. The constant $4a$ is not decorative: it falls straight out of the algebra of equating those two distances, and it is the reason the latus rectum length is always $|4a|$. This focus–directrix balance is also why parabolas are everywhere in the physical world — the path of a projectile under gravity, the cross-section of a satellite dish, the reflector behind a torch bulb or a vehicle headlight. A parabolic mirror gathers every ray parallel to the axis to the single focus, a reflective property that drops straight out of the equal-distance definition. Master the four forms not as four separate facts but as one idea seen from four directions: square one variable, set it equal to $4a$ times the other, and let the sign point the way.
Find the focus, directrix and length of the latus rectum of $y^2 = 12x$.
Solution- Compare with $y^2 = 4ax$: so $4a = 12$, giving $a = 3$.
- The parabola opens right, so the focus is $(a, 0) = (3, 0)$.
- Directrix is $x = -a$, i.e. $x = -3$.
- Length of latus rectum $= 4a = 12$.
Answer: Focus $(3, 0)$, directrix $x = -3$, latus rectum $12$.
Find the equation of the parabola with vertex at the origin, axis along the $y$-axis, passing through $(4, 2)$ and opening upward.
Solution- Axis along the $y$-axis and opening upward means the form $x^2 = 4ay$.
- Substitute the point $(4, 2)$: $4^2 = 4a(2) \Rightarrow 16 = 8a \Rightarrow a = 2$.
- So $4a = 8$ and the equation is $x^2 = 8y$.
Answer: $x^2 = 8y$.
Find the equation of the parabola with focus $(0, -3)$ and directrix $y = 3$.
Solution- The focus is below the origin and the directrix is above it, so the parabola opens downward: $x^2 = -4ay$.
- The focus is $(0, -a)$, so $-a = -3 \Rightarrow a = 3$.
- Thus $4a = 12$ and the equation is $x^2 = -12y$.
Answer: $x^2 = -12y$.
For the parabola $y^2 = -8x$, state which way it opens and give its focus.
Solution- Compare with $y^2 = -4ax$: so $4a = 8 \Rightarrow a = 2$.
- The negative sign with $y^2$ means it opens to the left.
- Focus of $y^2 = -4ax$ is $(-a, 0) = (-2, 0)$.
Answer: Opens left; focus $(-2, 0)$.
Find the equation of the parabola with vertex at the origin and focus $(5, 0)$.
Solution- The focus lies on the positive $x$-axis, so the parabola opens right: $y^2 = 4ax$.
- Focus is $(a, 0) = (5, 0)$, so $a = 5$.
- Then $4a = 20$, giving $y^2 = 20x$.
Answer: $y^2 = 20x$.
Find the coordinates of the endpoints of the latus rectum of $y^2 = 16x$.
Solution- Here $4a = 16 \Rightarrow a = 4$, so the focus is $(4, 0)$.
- The latus rectum is the vertical chord through the focus; substitute $x = 4$: $y^2 = 16(4) = 64 \Rightarrow y = \pm 8$.
- So the endpoints are $(4, 8)$ and $(4, -8)$ — a chord of length $16 = 4a$, as expected.
Answer: $(4, 8)$ and $(4, -8)$.
Find the focus, directrix and latus rectum of the parabola $x^2 = -16y$.
Solution- Compare with $x^2 = -4ay$: so $4a = 16 \Rightarrow a = 4$.
- The negative sign with $x^2$ means it opens downward, so the focus is $(0, -a) = (0, -4)$.
- Directrix is $y = a$, i.e. $y = 4$.
- Length of latus rectum $= 4a = 16$.
Answer: Focus $(0, -4)$, directrix $y = 4$, latus rectum $16$.
Find the equation of the parabola, vertex at the origin, whose axis is the $x$-axis and which passes through the point $(2, 3)$.
Solution- Axis along the $x$-axis and the point lies to the right, so use $y^2 = 4ax$.
- Substitute $(2, 3)$: $3^2 = 4a(2) \Rightarrow 9 = 8a \Rightarrow a = \dfrac{9}{8}$.
- Then $4a = \dfrac{9}{2}$, giving $y^2 = \dfrac{9}{2}x$, or $2y^2 = 9x$.
Answer: $y^2 = \dfrac{9}{2}x$ (equivalently $2y^2 = 9x$).
The focal distance of a point on the parabola $y^2 = 8x$ is $6$. Find the coordinates of that point.
Solution- Here $4a = 8 \Rightarrow a = 2$. For $y^2 = 4ax$ the focal distance of a point $(x, y)$ is $x + a = x + 2$.
- Set $x + 2 = 6 \Rightarrow x = 4$.
- Find $y$: $y^2 = 8(4) = 32 \Rightarrow y = \pm 4\sqrt{2}$.
Answer: $(4, 4\sqrt{2})$ and $(4, -4\sqrt{2})$.
A satellite-dish reflector is a parabola $40$ cm across at its rim and $10$ cm deep at the centre. How far from the vertex should the receiver (the focus) be placed?
Solution- Put the vertex at the origin with the dish opening along the positive $x$-axis: $y^2 = 4ax$.
- The rim is $40$ cm wide and $10$ cm deep, so the rim points are $(10, \pm 20)$.
- Substitute $(10, 20)$: $20^2 = 4a(10) \Rightarrow 400 = 40a \Rightarrow a = 10$.
- The focus is at distance $a$ from the vertex.
Answer: The receiver should be $10$ cm from the vertex (at the focus).
Find the equation of the parabola with focus $(-2, 0)$ and directrix $x = 2$.
Solution- The focus is on the negative $x$-axis and the directrix is the vertical line $x = 2$, so the parabola opens left: $y^2 = -4ax$.
- Focus is $(-a, 0) = (-2, 0)$, so $a = 2$.
- Then $4a = 8$, giving $y^2 = -8x$.
Answer: $y^2 = -8x$.
Find the vertex of the parabola $x = y^2 - 4y + 2$ by completing the square.
Solution- Group the $y$-terms and complete the square: $x = (y^2 - 4y + 4) - 4 + 2 = (y-2)^2 - 2$.
- Rearrange: $x + 2 = (y - 2)^2$, which is of the form $(y-k)^2 = (x-h)$ with $h = -2$, $k = 2$.
- The vertex is $(h, k)$.
Answer: Vertex $(-2, 2)$.