11.3 Unified Properties of Conic Sections and Trajectory Problems

11.3 Unified Properties of Conic Sections and Trajectory Problems

In high school mathematics and the Gaokao examination, conic sections are unified under a single geometric definition involving a focus, a directrix, and eccentricity $e$. Trajectory (locus) problems represent one of the most challenging problem types in analytic geometry, testing students' ability to translate geometric conditions into algebraic equations and manipulate quadratic systems.

---

1. The Unified Focus-Directrix Definition

Any non-degenerate conic section (ellipse, parabola, or hyperbola) can be defined as the locus of a point $P(x, y)$ moving in a plane such that the ratio of its distance from a fixed point $F$ (the focus) to its distance from a fixed line $l$ (the directrix, where $F \notin l$) is a constant non-negative real number $e$ (the eccentricity).

$$\frac{|PF|}{d(P, l)} = e$$

Unified Polar Equation: Setting the pole at the focus $F$ and polar axis along the axis of symmetry towards the directrix, the unified equation for all conic sections is: $r = \frac{ep}{1 - e \cos \theta}$

• When $e < 1$, the denominator $1 - e \cos \theta > 0$ for all $\theta$, giving a bounded closed ellipse.
• When $e = 1$, $1 - \cos \theta = 0$ at $\theta = 0$, giving an unbounded parabola.
• When $e > 1$, $1 - e \cos \theta = 0$ at $\cos \theta = \frac{1}{e}$, producing two asymptotic directions of a hyperbola.

---

2. Four Systematic Methods for Trajectory (Locus) Determination

Solving trajectory problems in the Gaokao requires identifying which of the four standard algebraic strategies applies to the geometric condition provided.

Method 1: Definition Method (定义法)

Strategy: Examine the geometric condition given in the problem statement. If it matches a known canonical definition of a conic section, directly identify the curve type and compute its parameters ($a, b, c,$ or $p$).

• Ellipse: $|PF_1| + |PF_2| = 2a > 2c$
• Hyperbola: $\bigl||PF_1| - |PF_2|\bigr| = 2a < 2c$
• Parabola: $|PF| = d(P, l)$

Method 2: Direct Algebraic Method (直译法)

Strategy: Set up a Cartesian coordinate system, assign coordinates $P(x, y)$ to the moving point, express the given geometric relations directly as an equation $f(x, y) = 0$, and simplify algebraically while taking care to state domain restrictions ($x$ or $y$ bounds).

Method 3: Related Point / Transfer Method (相关点法 / 代入法)

Strategy: When the trajectory of point $P(x,y)$ depends on the motion of another point $Q(x_0, y_0)$ that moves along a known curve $C: g(x_0, y_0) = 0$:

1. Express $x_0$ and $y_0$ in terms of $x$ and $y$: $x_0 = h_1(x, y)$ and $y_0 = h_2(x, y)$.
2. Substitute $(x_0, y_0)$ into the known curve equation $g(h_1(x,y), h_2(x,y)) = 0$.
3. Simplify to obtain the desired trajectory equation for $P(x, y)$.

Method 4: Parameter Elimination Method (参数法)

Strategy: Express the coordinates $P(x, y)$ of the moving point in terms of an auxiliary parameter $t$ (such as slope $k$, angle $\theta$, or time $t$):

$$\begin{cases} x = f(t) \\ y = g(t) \end{cases}$$

Eliminate $t$ between the two parametric equations to obtain the Cartesian equation $F(x, y) = 0$, ensuring the range of $t$ correctly translates into the domain/range of $x$ and $y$.

---

3. Advanced Algebraic Techniques for Line-Conic Intersections

When a line $l: y = kx + m$ (or $x = my + n$) intersects a conic section $C: Ax^2 + By^2 = 1$, Gaokao problems systematically test the Delta-Vieta System:

1. Substitution & Quadratic Reduction: Substitute $y = kx + m$ into the conic equation to obtain a standard single-variable quadratic equation: $$Ax^2 + B(kx + m)^2 = 1 \implies (A + B k^2)x^2 + 2BKmx + (Bm^2 - 1) = 0$$
2. Coefficient Non-Zero Check: Verify that the quadratic coefficient $A + Bk^2 \neq 0$.
   • Caution: If $A + Bk^2 = 0$, the line is parallel to an asymptote of a hyperbola, resulting in at most one intersection point!
3. Discriminant Constraint ($\Delta > 0$): Compute $\Delta = (2BKm)^2 - 4(A + Bk^2)(Bm^2 - 1) > 0$ to guarantee two real distinct intersection points $M(x_1, y_1)$ and $N(x_2, y_2)$.
4. Vieta's Formulas: Express $x_1 + x_2$ and $x_1 x_2$ in terms of parameters $k$ and $m$.
5. Point-Subtraction Method (点差法) for Midpoint Chords: If $M(x_1, y_1)$ and $N(x_2, y_2)$ lie on $\frac{x^2}{a^2} \pm \frac{y^2}{b^2} = 1$ with chord midpoint $P_0(x_0, y_0)$: $$\frac{x_1^2 - x_2^2}{a^2} \pm \frac{y_1^2 - y_2^2}{b^2} = 0 \implies k_{\text{chord}} \cdot k_{OP_0} = \mp \frac{b^2}{a^2}$$

---

4. Worked Gaokao Exam Examples

Example 1: Trajectory of a Moving Circle Center (Definition Method)

Problem: A moving circle $M$ is externally tangent to circle $C_1: (x+2)^2 + y^2 = 1$ and internally tangent to circle $C_2: (x-2)^2 + y^2 = 25$. Find the trajectory equation of the center $M(x, y)$ of the moving circle.

Solution:

1. Circle $C_1$ has center $F_1(-2, 0)$ and radius $R_1 = 1$. Circle $C_2$ has center $F_2(2, 0)$ and radius $R_2 = 5$.
2. Let $r$ be the radius of the moving circle $M$.
3. External tangency to $C_1 \implies |MF_1| = r + R_1 = r + 1$.
4. Internal tangency to $C_2 \implies |MF_2| = R_2 - r = 5 - r$.
5. Adding the two distance equations: $$|MF_1| + |MF_2| = (r + 1) + (5 - r) = 6$$
6. Notice that $|F_1F_2| = 4$. Since $|MF_1| + |MF_2| = 6 > |F_1F_2| = 4$, by the definition of an ellipse, the locus of $M$ is an ellipse with foci $F_1(-2, 0)$ and $F_2(2, 0)$.
7. Here $2a = 6 \implies a = 3$, and $c = 2$. Thus $b^2 = a^2 - c^2 = 9 - 4 = 5$.
8. The trajectory equation of $M$ is $\dfrac{x^2}{9} + \dfrac{y^2}{5} = 1$.

---

Example 2: Trajectory by Related Point Method

Problem: A point $Q(x_0, y_0)$ moves along the parabola $y^2 = 4x$. Point $P(x, y)$ divides the segment $OQ$ (where $O$ is the origin) such that $\vec{OP} = \frac{2}{3}\vec{OQ}$. Find the equation of the trajectory of $P$.

Solution:

1. Express the vector relation in coordinates: $$(x, y) = \frac{2}{3}(x_0, y_0) \implies x = \frac{2}{3}x_0, \quad y = \frac{2}{3}y_0$$
2. Solve for $x_0$ and $y_0$ in terms of $x$ and $y$: $$x_0 = \frac{3}{2}x, \quad y_0 = \frac{3}{2}y$$
3. Since $Q(x_0, y_0)$ lies on $y^2 = 4x$, substitute $(x_0, y_0)$ into $y_0^2 = 4x_0$: $$\left(\frac{3}{2}y\right)^2 = 4\left(\frac{3}{2}x\right) \implies \frac{9}{4}y^2 = 6x$$
4. Multiply both sides by $\frac{4}{9}$: $$y^2 = \frac{24}{9}x = \frac{8}{3}x$$
5. The trajectory of point $P$ is the parabola $y^2 = \dfrac{8}{3}x$.