10.1 Standard Equations and Definitions of Ellipses

10.1 Standard Equations and Definitions of Ellipses

The ellipse is one of the foundational conic sections heavily tested in the Gaokao (Chinese National College Entrance Examination) Mathematics exam. A deep understanding of its geometric definition, algebraic derivation, and standard forms is essential for solving both single-choice questions and complex analytic geometry synthesis problems (解析几何综合题).

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1. Geometric Definition of an Ellipse

First Definition (Locus Definition)

An ellipse is defined as the locus of a point $P(x, y)$ in a two-dimensional plane such that the sum of its distances to two fixed points $F_1$ and $F_2$ (called the foci) remains constant.

$|PF_1| + |PF_2| = 2a$

where $2a > 2c = |F_1F_2| > 0$.

Here, $a$ represents the semi-major axis length, and $c$ represents the half-focal distance (distance from the center to either focus).

Critical Boundary Conditions & Degenerate Cases

Understanding the relationship between $2a$ and $2c$ is a frequent trap in Gaokao conceptual questions:

1. When $2a > 2c > 0$: The locus of point $P$ is a non-degenerate ellipse.
2. When $2a = 2c$: The locus of point $P$ degrades into the straight line segment connecting $F_1$ and $F_2$ ($[F_1F_2]$).
3. When $2a < 2c$: No such point $P$ exists in the real Cartesian plane (the locus is the empty set $\emptyset$).

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2. Derivation of Standard Equations

To derive the standard equation of an ellipse centered at the origin $(0,0)$, we establish a rectangular Cartesian coordinate system where the line passing through foci $F_1$ and $F_2$ serves as one coordinate axis.

Case 1: Foci on the x-axis ($F_1(-c, 0)$ and $F_2(c, 0)$)

Let $P(x, y)$ be any point on the ellipse. By the locus definition: $\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a$

Step-by-Step Algebraic Derivation:

1. Isolate one radical: $\sqrt{(x+c)^2 + y^2} = 2a - \sqrt{(x-c)^2 + y^2}$

2. Square both sides: $(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} + (x-c)^2 + y^2$

3. Expand and simplify terms: $x^2 + 2cx + c^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} + x^2 - 2cx + c^2 + y^2$ $4cx = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} \implies cx - a^2 = -a\sqrt{(x-c)^2 + y^2}$

4. Square both sides again: $(a^2 - cx)^2 = a^2 \left( (x-c)^2 + y^2 \right)$ $a^4 - 2a^2cx + c^2x^2 = a^2(x^2 - 2cx + c^2 + y^2)$ $a^4 - 2a^2cx + c^2x^2 = a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2$

5. Rearrange terms: $(a^2 - c^2)x^2 + a^2y^2 = a^2(a^2 - c^2)$

6. Introduce parameter $b$: Since $a > c > 0$, we have $a^2 - c^2 > 0$. We define the semi-minor axis length $b$ such that: $b^2 = a^2 - c^2 \quad (b > 0)$

7. Divide by $a^2b^2$: $b^2x^2 + a^2y^2 = a^2b^2 \implies \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b > 0)$

Case 2: Foci on the y-axis ($F_1(0, -c)$ and $F_2(0, c)$)

When the foci lie along the vertical y-axis, swapping the roles of $x$ and $y$ yields: $\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \quad (a > b > 0)$

Gaokao Rule of Thumb: The larger denominator always corresponds to the variable associated with the axis containing the foci! If the denominator under $x^2$ is larger, the foci lie on the x-axis. If the denominator under $y^2$ is larger, the foci lie on the y-axis.

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3. General & Translated Forms of Ellipse Equations

For the general quadratic equation $Ax^2 + Cy^2 + Dx + Ey + F = 0$ to represent an ellipse:

• $A$ and $C$ must have the same sign ($AC > 0$).
• $A \neq C$ (if $A = C$, the curve represents a circle).
• After completing the square, the constant right-hand side term must be strictly positive: $\frac{D^2}{4A} + \frac{E^2}{4C} - F > 0$.

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4. Parametric Equations of an Ellipse

For the standard horizontal ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, introducing an auxiliary parameter $\theta$ (eccentric angle, 离心角) gives the parametric representation:

$\begin{cases} x = a \cos\theta \\ y = b \sin\theta \end{cases} \quad (\theta \in [0, 2\pi))$

Geometric Meaning of the Eccentric Angle $\theta$

Consider two concentric circles centered at the origin:

1. Outer circle (Major auxiliary circle): $x^2 + y^2 = a^2$.
2. Inner circle (Minor auxiliary circle): $x^2 + y^2 = b^2$.

A ray originating from the origin at angle $\theta$ with the positive x-axis intersects the outer circle at $M(a\cos\theta, a\sin\theta)$ and the inner circle at $N(b\cos\theta, b\sin\theta)$. The point $P(a\cos\theta, b\sin\theta)$ obtained by projecting $M$ vertically and $N$ horizontally lies precisely on the ellipse!

Exam Application: Parametric forms simplify optimization problems on ellipses (such as finding the maximum distance from a point to a line or maximizing linear combinations $Ax + By$) into single-variable trigonometric extremum problems.

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5. Worked Gaokao Exam Examples

Example 1: Determining Standard Equation from General Form

Problem: Given the equation $3x^2 + 4y^2 - 12x + 8y - 4 = 0$, verify that it represents an ellipse, determine its center $(h,k)$, semi-major axis $a$, semi-minor axis $b$, focal distance $c$, and foci coordinates.

Solution:

1. Group and complete the square for $x$ and $y$: $3(x^2 - 4x) + 4(y^2 + 2y) = 4$ $3(x - 2)^2 - 12 + 4(y + 1)^2 - 4 = 4$ $3(x - 2)^2 + 4(y + 1)^2 = 20$

2. Convert to standard form by dividing by 20: $\frac{(x - 2)^2}{\frac{20}{3}} + \frac{(y + 1)^2}{5} = 1$

3. Extract parameters:

   • Compare denominators: $a^2 = \frac{20}{3} \approx 6.67$, $b^2 = 5$. Since $\frac{20}{3} > 5$, the major axis is horizontal.
   • Semi-major axis: $a = \sqrt{\frac{20}{3}} = \frac{2\sqrt{15}}{3}$.
   • Semi-minor axis: $b = \sqrt{5}$.
   • Center: $(h,k) = (2, -1)$.
   • Half-focal distance: $c = \sqrt{a^2 - b^2} = \sqrt{\frac{20}{3} - 5} = \sqrt{\frac{5}{3}} = \frac{\sqrt{15}}{3}$.
   • Foci coordinates: $(h \pm c, k) = \left(2 \pm \frac{\sqrt{15}}{3}, -1\right)$.

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Example 2: Determining Locus Equation under Geometric Conditions (Gaokao Standard Problem)

Problem: In $\triangle ABC$, the base vertices are $B(-3,0)$ and $C(3,0)$. If the perimeter of $\triangle ABC$ is 16, find the trajectory equation of vertex $A$.

Solution:

1. Analyze perimeter condition: The perimeter of $\triangle ABC$ is $|AB| + |AC| + |BC| = 16$. The length of base $BC$ is $|BC| = \sqrt{(3 - (-3))^2 + 0^2} = 6$. Therefore, $|AB| + |AC| = 16 - 6 = 10$.

2. Apply ellipse locus definition: The sum of distances from moving point $A(x,y)$ to fixed points $B(-3,0)$ and $C(3,0)$ is constant $2a = 10$. Here $c = 3$. Since $2a = 10 > 2c = 6$, the locus of $A$ is an ellipse with foci at $B$ and $C$.

3. Calculate parameters: $a = 5$, $c = 3 \implies b^2 = a^2 - c^2 = 25 - 9 = 16 \implies b = 4$.

4. Account for degenerate triangle constraint: Since $A, B, C$ must form a triangle, $A$ cannot lie on the x-axis (i.e., $A \neq (-5,0)$ and $A \neq (5,0)$). Thus $y \neq 0$.

5. Final Equation: $\frac{x^2}{25} + \frac{y^2}{16} = 1 \quad (y \neq 0 \text{ or } x \neq \pm 5)$