Find the equation of the parabola with focus $(2, 0)$ and directrix $x = -2$.

Question

Accepted Answer

> [!TLDR]
> The parabola has its focus at $(2,0)$ and directrix $x=-2$, so the vertex is at the origin and the axis is horizontal. The equation is $y^2 = 8x$.

The definition of a parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). This distance condition is what gives us the equation directly — no memorised formulas needed, just the distance formula and algebra.

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1. **Set up the distance condition.**  
   Let $P(x, y)$ be any point on the parabola.  
   Distance from $P$ to the focus $F(2, 0)$:  
   $$PF = \sqrt{(x-2)^2 + (y-0)^2}$$  
   Distance from $P$ to the directrix $x = -2$: the perpendicular distance from a point to a vertical line is the absolute horizontal difference:  
   $$PD = |x - (-2)| = |x+2|$$

2. **Equate the two distances.**  
   By definition, $PF = PD$:  
   $$\sqrt{(x-2)^2 + y^2} = |x+2|$$

3. **Square both sides** (both sides are non-negative, so no sign issues):  
   $$(x-2)^2 + y^2 = (x+2)^2$$

4. **Expand and simplify.**  
   $$x^2 - 4x + 4 + y^2 = x^2 + 4x + 4$$  
   Cancel $x^2$ and $4$ from both sides:  
   $$-4x + y^2 = 4x$$  
   $$y^2 = 8x$$

> [!TIP]
> Notice the $x^2$ terms cancel immediately — that’s a sign the parabola opens sideways. If the directrix were horizontal, the $y^2$ terms would cancel instead.

5. **Interpret the result.**  
   The equation $y^2 = 8x$ is in the standard form $y^2 = 4ax$, where $4a = 8$, so $a = 2$.  
   The vertex is at $(0,0)$, the focus is at $(a,0) = (2,0)$, and the directrix is $x = -a = -2$ — which matches the given data perfectly.

> [!WARNING]
> A common mistake is to write $y^2 = -8x$ or $x^2 = 8y$ by mixing up which variable is squared. The focus $(2,0)$ lies on the $x$-axis, so the parabola opens to the right — $y$ is squared, $x$ is linear.

> [!ANSWER]
> The equation of the parabola is $\boxed{y^2 = 8x}$.

Equation	Opens	Focus	Directrix
$y^2 = 4ax$	right	$(a, 0)$	$x = -a$
$y^2 = -4ax$	left	$(-a, 0)$	$x = a$
$x^2 = 4ay$	up	$(0, a)$	$y = -a$
$x^2 = -4ay$	down	$(0, -a)$	$y = a$

Q.Find the equation of the parabola with focus $(2, 0)$ and directrix $x = -2$ .

Concept understanding — Parabola Standard Form

Standard Equations of a Parabola

The four standard forms

Where $y^2 = 4ax$ comes from

Latus rectum

Worked example

Q.Find the equation of the parabola with focus (2,0)(2, 0)(2,0) and directrix x=−2x = -2x=−2.

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Q.Find the equation of the parabola with focus $(2, 0)$ and directrix $x = -2$ .

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