10.2 Geometric Properties of Ellipses (Vertices, Foci, Eccentricity)

10.2 Geometric Properties of Ellipses (Vertices, Foci, Eccentricity)

In the Gaokao Mathematics syllabus, mastering the intrinsic geometric properties of ellipses—including vertices, symmetry, eccentricity, focal radii, and focal triangles—is crucial. Questions in this section range from quick identification in Section I choices to deep algebraic derivations in Section II comprehensive problems.

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1. Fundamental Parameters and Symmetry Properties

For the standard horizontal ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$):

Basic Geometric Dimensions

• Semi-major axis: $a$ (length of major axis $= 2a$).
• Semi-minor axis: $b$ (length of minor axis $= 2b$).
• Focal distance: $2c$, where $c = \sqrt{a^2 - b^2}$.
• Fundamental Identity: $$a^2 = b^2 + c^2$$

Vertices and Center

• Principal Vertices (on major axis): $A_1(-a, 0)$ and $A_2(a, 0)$.
• Co-vertices (on minor axis): $B_1(0, -b)$ and $B_2(0, b)$.
• Center: The origin $(0,0)$, which is the midpoint of both major and minor axes.

Symmetry

• Axes of Symmetry: The x-axis ($y=0$) and y-axis ($x=0$). Replacing $x$ with $-x$ or $y$ with $-y$ leaves the equation invariant.
• Center of Symmetry: The origin $(0,0)$. Replacing $(x,y)$ with $(-x,-y)$ leaves the equation invariant.

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2. Eccentricity ($e$) and Geometric Significance

Definition and Range

The eccentricity $e$ measures the degree of flatness of an ellipse:

$$\ne = \frac{c}{a} = \sqrt{1 - \left(\frac{b}{a}\right)^2} \quad (0 < e < 1)$$

Geometric Behavior as $e$ Varies:

1. As $e \to 0^+$: $c \to 0$ and $b \to a$. The foci approach the center, and the ellipse approaches a circle.
2. As $e \to 1^-$: $c \to a$ and $b \to 0$. The ellipse flattens out, approaching the major axis line segment $[-a, a]$.

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3. Second Definition of an Ellipse & Directrices

The Second Definition (Focal Ratio Definition)

An ellipse is the locus of points $P(x,y)$ such that the ratio of the distance from $P$ to a focus $F$ to the distance from $P$ to a fixed vertical line $d$ (called the directrix) is a constant equal to the eccentricity $e$:

$$\frac{|PF|}{\text{dist}(P, d)} = e = \frac{c}{a} < 1$$

Directrix Equations

For the horizontal ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

• Left Directrix $l_1$: $x = -\frac{a^2}{c}$
• Right Directrix $l_2$: $x = \frac{a^2}{c}$

For the vertical ellipse $\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1$:

• Bottom Directrix: $y = -\frac{a^2}{c}$
• Top Directrix: $y = \frac{a^2}{c}$

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4. Focal Radii Formulas (焦半径公式)

Let $P(x_0, y_0)$ be any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$). The distances from $P$ to the left focus $F_1(-c, 0)$ and right focus $F_2(c, 0)$ are known as the focal radii:

$$|PF_1| = a + ex_0, \quad |PF_2| = a - ex_0$$

Proof:

Using the second definition with the right directrix $x = \frac{a^2}{c}$:

$$\text{dist}(P, l_2) = \frac{a^2}{c} - x_0$$ $$|PF_2| = e \cdot \text{dist}(P, l_2) = \frac{c}{a} \left( \frac{a^2}{c} - x_0 \right) = a - ex_0$$

Similarly, for the left focus:

$$|PF_1| = 2a - |PF_2| = 2a - (a - ex_0) = a + ex_0$$

Extremum Bounds of Focal Radii:

Since $-a \le x_0 \le a$:

• Minimum focal radius: $|PF|_{\min} = a - c$ (achieved at vertex closest to the focus).
• Maximum focal radius: $|PF|_{\max} = a + c$ (achieved at vertex furthest from the focus).

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5. Focal Triangles (焦点三角形) and Area Formula

A focal triangle $\triangle F_1 P F_2$ is formed by connecting any point $P(x_0, y_0)$ on the ellipse to the two foci $F_1$ and $F_2$.

The Focal Triangle Area Formula

Let $\theta = \angle F_1 P F_2$ be the angle subtended at point $P$. The area of $\triangle F_1 P F_2$ is given by:

$$S_{\triangle F_1 P F_2} = b^2 \tan\left(\frac{\theta}{2}\right)$$

Rigorous Proof:

1. In $\triangle F_1 P F_2$, by the Law of Cosines:

   $$|F_1F_2|^2 = |PF_1|^2 + |PF_2|^2 - 2|PF_1||PF_2|\cos\theta$$ $$(2c)^2 = (|PF_1| + |PF_2|)^2 - 2|PF_1||PF_2| - 2|PF_1||PF_2|\cos\theta$$ $$4c^2 = (2a)^2 - 2|PF_1||PF_2|(1 + \cos\theta)$$

2. Rearrange to find $2|PF_1||PF_2|(1 + \cos\theta)$:

   $$2|PF_1||PF_2|(1 + \cos\theta) = 4a^2 - 4c^2 = 4b^2$$ $$|PF_1||PF_2| = \frac{2b^2}{1 + \cos\theta}$$

3. Express Area $S$:

   $$S = \frac{1}{2} |PF_1| |PF_2| \sin\theta = \frac{1}{2} \left( \frac{2b^2}{1 + \cos\theta} \right) \sin\theta = b^2 \frac{\sin\theta}{1 + \cos\theta}$$

4. Use the trigonometric half-angle identity $\frac{\sin\theta}{1 + \cos\theta} = \tan\left(\frac{\theta}{2}\right)$:

   $$S_{\triangle F_1 P F_2} = b^2 \tan\left(\frac{\theta}{2}\right)$$

Maximum Area: Since $S = \frac{1}{2} |F_1F_2| |y_0| = c |y_0|$, the maximum area occurs when $|y_0| = b$ (at co-vertices $B_1, B_2$): $S_{\max} = bc$

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6. Latus Rectum (通径)

The latus rectum of an ellipse is the chord passing through either focus perpendicular to the major axis.

Substituting $x = c$ into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$$\frac{c^2}{a^2} + \frac{y^2}{b^2} = 1 \implies \frac{y^2}{b^2} = 1 - \frac{c^2}{a^2} = \frac{b^2}{a^2} \implies y = \pm \frac{b^2}{a}$$

Thus, the length of the latus rectum $d$ is:

$$d = \frac{2b^2}{a}$$

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7. Worked Gaokao Exam Examples

Example 1: Finding Eccentricity from Geometric Right Angle Condition

Problem: For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$), if the angle between the lines connecting the top co-vertex $B(0,b)$ to the two foci $F_1(-c,0)$ and $F_2(c,0)$ is a right angle ($\angle F_1 B F_2 = 90^\circ$), find the eccentricity $e$.

Solution:

1. Analyze vector/geometric relation: In right-angled isosceles $\triangle F_1 B F_2$, the height $OB = b$ and half-base $OF_2 = c$. Since $\angle F_1 B F_2 = 90^\circ$, $\angle O B F_2 = 45^\circ$, so $b = c$.

2. Substitute into fundamental identity:

   $$a^2 = b^2 + c^2 = c^2 + c^2 = 2c^2$$

3. Calculate eccentricity:

   $$e = \frac{c}{a} = \frac{c}{\sqrt{2}c} = \frac{\sqrt{2}}{2}$$

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Example 2: Focal Radii Optimization in Gaokao Synthesis

Problem: Given ellipse $C: \frac{x^2}{4} + y^2 = 1$ with foci $F_1, F_2$. Point $P$ is on the ellipse, and point $A(1, 1)$ is fixed in the plane. Find the minimum value of $|PA| + |PF_1|$.

Solution:

1. Identify parameters: $a^2 = 4 \implies a = 2$, $b^2 = 1$, $c = \sqrt{4-1} = \sqrt{3}$. Foci are $F_1(-\sqrt{3}, 0)$ and $F_2(\sqrt{3}, 0)$.

2. Use ellipse definition to replace $|PF_1|$: By definition, $|PF_1| + |PF_2| = 2a = 4 \implies |PF_1| = 4 - |PF_2|$. Therefore:

   $$|PA| + |PF_1| = |PA| + 4 - |PF_2| = 4 + (|PA| - |PF_2|)$$

3. Apply Triangle Inequality: In $\triangle A P F_2$, by the triangle inequality:

   $$|PA| - |PF_2| \ge -|AF_2|$$

   The minimum occurs when $P, A, F_2$ are collinear in the order $P - A - F_2$ (i.e., $P$ lies on the ray $F_2A$).

4. Calculate distance $|AF_2|$:

   $$|AF_2| = \sqrt{(1 - \sqrt{3})^2 + (1 - 0)^2} = \sqrt{1 - 2\sqrt{3} + 3 + 1} = \sqrt{5 - 2\sqrt{3}}$$

5. Final Minimum Value:

   $$\min (|PA| + |PF_1|) = 4 - |AF_2| = 4 - \sqrt{5 - 2\sqrt{3}}$$