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Equations and Properties of Parabolas

CSCA Equations and Properties of Parabolas study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

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Equations and Properties of Parabolas

The parabola is the only conic section with an "eccentricity $e=1$". The key to this topic is applying the **geometric definition**. In the CSCA exam, mastering the four standard equations and using the "Definition Method" are crucial.

1. Definition

A parabola is the locus of points in a plane that are equidistant from a fixed point $F$ (**Focus**) and a fixed line $l$ (**Directrix**).

* **Condition**: $F \notin l$ (The focus is not on the directrix).

* **Focal Parameter $p$**: The distance from the focus $F$ to the directrix $l$, i.e., $d(F, l) = p$ ($p > 0$).

* **Geometric Formula**: Let $P$ be any point on the parabola, and $d$ be the distance from $P$ to the directrix. Then **$|PF| = d$**.

2. Standard Equations

Establish a coordinate system: The axis of symmetry passes through the focus and is perpendicular to the directrix. The origin is the midpoint between the focus and the directrix.

There are four standard forms depending on the opening direction. **Memory Rule**: "The variable in the quadratic term determines the perpendicular axis; the variable in the linear term determines the axis of symmetry."

| Direction | **Right** | **Left** | **Up** | **Down** |

| :--- | :---: | :---: | :---: | :---: |

| **Graph** | | (Opens Left) | (Opens Up) | (Opens Down) |

| **Equation** | $$y^2 = 2px$$ | $$y^2 = -2px$$ | $$x^2 = 2py$$ | $$x^2 = -2py$$ |

| **Focus** | $$(\frac{p}{2}, 0)$$ | $$(-\frac{p}{2}, 0)$$ | $$(0, \frac{p}{2})$$ | $$(0, -\frac{p}{2})$$ |

| **Directrix** | $$x = -\frac{p}{2}$$ | $$x = \frac{p}{2}$$ | $$y = -\frac{p}{2}$$ | $$y = \frac{p}{2}$$ |

3. Geometric Properties

Taking the right-opening parabola $y^2 = 2px$ as an example:

1. **Range**: $x \ge 0$, $y \in \mathbb{R}$.

2. **Symmetry**: Symmetric about the x-axis.

3. **Vertex**: The origin $O(0,0)$.

4. **Eccentricity**: $e = 1$.

5. **Focal Radius**: The distance from a point $P(x_0, y_0)$ on the parabola to the focus $F$.

* Formula: $|PF| = x_0 + \frac{p}{2}$ (Derived from definition: $|PF| = d = x_0 - (-\frac{p}{2})$).

6. **Latus Rectum**: The chord passing through the focus and perpendicular to the axis of symmetry. Its length is **$2p$**.

4. Examples

**Example 1**: Find the equation of the parabola with focus $F(3, 0)$.

**Solution**:

Focus on positive x-axis $\Rightarrow$ Opens Right, form is $y^2 = 2px$.

From focus $(\frac{p}{2}, 0) = (3, 0)$, we get $\frac{p}{2} = 3 \Rightarrow p = 6$.

Equation: $y^2 = 2(6)x \Rightarrow y^2 = 12x$.

**Example 2**: A parabola passes through $A(2, -4)$, vertex at origin, axis along coordinate axes. Find the equation.

**Solution**:

Point $A(2, -4)$ is in Q4.

* **Case 1**: Symmetric about x-axis (Opens Right since $x>0$).

Let $y^2 = 2px$. Substitute $A$: $(-4)^2 = 2p(2) \Rightarrow 16 = 4p \Rightarrow p = 4$.

Equation: $y^2 = 8x$.

* **Case 2**: Symmetric about y-axis (Opens Down since $y<0$).

Let $x^2 = -2py$. Substitute $A$: $2^2 = -2p(-4) \Rightarrow 4 = 8p \Rightarrow p = 0.5$.

Equation: $x^2 = -y$.

**Conclusion**: $y^2 = 8x$ or $x^2 = -y$.

**Example 3**: A point $P$ on $y^2 = 8x$ is at a distance of 5 from the focus. Find the x-coordinate of $P$.

**Solution**:

From equation, $2p = 8 \Rightarrow p = 4$.

Using Focal Radius formula: $|PF| = x_P + \frac{p}{2}$.

$5 = x_P + \frac{4}{2} \Rightarrow 5 = x_P + 2 \Rightarrow x_P = 3$.

5. Study Tips

* **Meaning of $p$**: Remember $p$ is "distance from Focus to Directrix", not Focus to Origin (which is $p/2$).

* **Definition Method**: When you see "distance to focus", immediately think "distance to directrix".

* **Check Direction**: Look at the linear variable. Linear $x$ $\to$ Left/Right; Linear $y$ $\to$ Up/Down.