10.3 Intersection of Lines and Ellipses and Chord Problems

10.3 Intersection of Lines and Ellipses and Chord Problems

Line-ellipse interaction problems constitute the single most tested topic in the comprehensive analytic geometry question (解析几何大题) on the Gaokao exam. These problems test algebraic manipulation, discriminant analysis, Vieta's formulas, chord length formulas, and the point-difference method (点差法).

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1. Algebraic System & Intersection Discriminant ($\Delta$)

To investigate the intersection of a line $l: y = kx + m$ and an ellipse $C: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

Step 1: Substitute and Form the Quadratic System

Substitute $y = kx + m$ into the ellipse equation: $\frac{x^2}{a^2} + \frac{(kx + m)^2}{b^2} = 1$ Multiply by $a^2b^2$: $b^2 x^2 + a^2(k^2 x^2 + 2kmx + m^2) = a^2b^2$ $(b^2 + a^2 k^2) x^2 + 2a^2 k m x + a^2(m^2 - b^2) = 0$

Let $A = b^2 + a^2 k^2$, $B = 2a^2 k m$, and $C = a^2(m^2 - b^2)$.

Step 2: Compute Discriminant $\Delta$

$\Delta = B^2 - 4AC = (2a^2 k m)^2 - 4(b^2 + a^2 k^2) a^2 (m^2 - b^2)$ $\Delta = 4a^4 k^2 m^2 - 4a^2 (b^2 m^2 - b^4 + a^2 k^2 m^2 - a^2 k^2 b^2)$ $\Delta = 4a^2 b^2 (a^2 k^2 + b^2 - m^2)$

Classification of Geometric States:

1. $\Delta > 0 \iff m^2 < a^2 k^2 + b^2$: Line $l$ intersects ellipse $C$ at two distinct points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ (secant line).
2. $\Delta = 0 \iff m^2 = a^2 k^2 + b^2$: Line $l$ is tangent to ellipse $C$ at exactly one point.
3. $\Delta < 0 \iff m^2 > a^2 k^2 + b^2$: Line $l$ and ellipse $C$ have no intersection points.

Tangency Condition: The condition for line $y = kx + m$ to be tangent to ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $m^2 = a^2k^2 + b^2$.

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2. Vieta's Formulas & Chord Length Formula

When line $y = kx + m$ intersects ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$:

Vieta's Formulas (韦达定理)

$x_1 + x_2 = -\frac{2a^2 km}{b^2 + a^2 k^2}, \quad x_1 x_2 = \frac{a^2(m^2 - b^2)}{b^2 + a^2 k^2}$

Chord Length Formula (弦长公式)

Using the distance formula $|P_1 P_2| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$: Since $y_1 - y_2 = k(x_1 - x_2)$, we substitute to obtain:

$|P_1 P_2| = \sqrt{1 + k^2} |x_1 - x_2| = \sqrt{1 + k^2} \sqrt{(x_1 + x_2)^2 - 4x_1 x_2}$

Alternative Vertical Parameterization:

If the line is expressed as $x = m y + n$ (useful when the line is vertical or $k \to \infty$): $|P_1 P_2| = \sqrt{1 + m^2} |y_1 - y_2| = \sqrt{1 + m^2} \sqrt{(y_1 + y_2)^2 - 4y_1 y_2}$

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3. The Point-Difference Method (点差法) for Midpoint Chords

For midpoint chord problems—where a chord $AB$ of an ellipse has a known midpoint $M(x_0, y_0)$—the point-difference method circumvents quadratic discriminant calculations.

Derivation:

Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be endpoints of chord $AB$, and $M(x_0, y_0)$ be its midpoint.

1. Since $A$ and $B$ lie on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1 \quad \text{and} \quad \frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} = 1$

2. Subtract the two equations: $\frac{x_1^2 - x_2^2}{a^2} + \frac{y_1^2 - y_2^2}{b^2} = 0$ $\frac{(x_1 - x_2)(x_1 + x_2)}{a^2} + \frac{(y_1 - y_2)(y_1 + y_2)}{b^2} = 0$

3. Substitute $x_1 + x_2 = 2x_0$, $y_1 + y_2 = 2y_0$, $y_1 - y_2 = k_{AB}(x_1 - x_2)$: $\frac{2x_0 (x_1 - x_2)}{a^2} + \frac{2y_0 (y_1 - y_2)}{b^2} = 0$ $\frac{x_0}{a^2} + \frac{y_0}{b^2} \cdot k_{AB} = 0 \implies k_{AB} = -\frac{b^2 x_0}{a^2 y_0}$

Key Slope Product Property:

Defining the slope from the origin to midpoint $M$ as $k_{OM} = \frac{y_0}{x_0}$:

$k_{AB} \cdot k_{OM} = -\frac{b^2}{a^2}$

Gaokao Theorem: For any chord $AB$ of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with midpoint $M$, the product of the slope of chord $AB$ ($k_{AB}$) and the slope of line $OM$ ($k_{OM}$) is a constant equal to $-\frac{b^2}{a^2}$.

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4. Worked Gaokao Exam Examples

Example 1: Midpoint Chord Determination via Point-Difference Method

Problem: Given the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$, find the equation of the line passing through point $M(2, 1)$ such that $M$ is the midpoint of the chord $AB$ intercepted by the ellipse.

Solution:

1. Apply the Slope Product Property: Here $a^2 = 16$ and $b^2 = 4$. Point $M(x_0, y_0) = (2, 1)$. Slope of line $OM$: $k_{OM} = \frac{y_0}{x_0} = \frac{1}{2}$. Using $k_{AB} \cdot k_{OM} = -\frac{b^2}{a^2}$: $k_{AB} \cdot \left(\frac{1}{2}\right) = -\frac{4}{16} = -\frac{1}{4} \implies k_{AB} = -\frac{1}{2}$

2. Write line equation using point-slope form: $y - 1 = -\frac{1}{2}(x - 2) \implies y = -\frac{1}{2}x + 2 \iff x + 2y - 4 = 0$

3. Verify Discriminant ($\Delta > 0$): Substitute $x = 4 - 2y$ into $\frac{x^2}{16} + \frac{y^2}{4} = 1$: $\frac{(4 - 2y)^2}{16} + \frac{y^2}{4} = 1 \implies \frac{16 - 16y + 4y^2}{16} + \frac{y^2}{4} = 1$ $1 - y + \frac{y^2}{4} + \frac{y^2}{4} = 1 \implies \frac{y^2}{2} - y = 0 \implies y(y - 2) = 0$ The roots are $y_1 = 0, y_2 = 2$. Midpoint $y_0 = \frac{0+2}{2} = 1$, which confirms $M(2,1)$ is inside the ellipse and is indeed the midpoint!

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Example 2: Gaokao Comprehensive Analytic Geometry Problem (解答题)

Problem: An ellipse $C: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$) passes through point $P(0, 1)$ and has eccentricity $e = \frac{\sqrt{3}}{2}$.

1. Find the standard equation of ellipse $C$.
2. A line $l: y = kx + m$ passes through point $T(0, 2)$ and intersects ellipse $C$ at two distinct points $A$ and $B$. If $\vec{OA} \cdot \vec{OB} = 0$ (where $O$ is the origin), find the value of $k$.

Solution Part 1: Determine Ellipse Equation:

1. Since $P(0, 1)$ is on the ellipse and lies on the y-axis, $b = 1$.
2. Eccentricity $e = \frac{c}{a} = \frac{\sqrt{3}}{2} \implies e^2 = \frac{c^2}{a^2} = \frac{a^2 - b^2}{a^2} = \frac{a^2 - 1}{a^2} = \frac{3}{4}$.
3. Solve for $a^2$: $4a^2 - 4 = 3a^2 \implies a^2 = 4$.
4. Standard equation of ellipse $C$: $\frac{x^2}{4} + y^2 = 1$

Solution Part 2: Line Intersections & Vector Condition:

1. Set up line equation: Since line $l$ passes through $T(0, 2)$, its equation is $y = kx + 2$.
2. Form quadratic equation: Substitute $y = kx + 2$ into $\frac{x^2}{4} + y^2 = 1$: $x^2 + 4(kx + 2)^2 = 4 \implies (1 + 4k^2)x^2 + 16kx + 12 = 0$
3. Discriminant condition: $\Delta = (16k)^2 - 4(1 + 4k^2)(12) = 256k^2 - 48 - 192k^2 = 64k^2 - 48 > 0 \implies k^2 > \frac{3}{4}$
4. Vieta's Formulas: $x_1 + x_2 = -\frac{16k}{1 + 4k^2}, \quad x_1 x_2 = \frac{12}{1 + 4k^2}$
5. Express Vector Dot Product $\vec{OA} \cdot \vec{OB}$: $\vec{OA} \cdot \vec{OB} = x_1 x_2 + y_1 y_2 = x_1 x_2 + (kx_1 + 2)(kx_2 + 2)$ $\vec{OA} \cdot \vec{OB} = (1 + k^2) x_1 x_2 + 2k(x_1 + x_2) + 4$
6. Substitute Vieta's relations into dot product: $\vec{OA} \cdot \vec{OB} = (1 + k^2) \left(\frac{12}{1 + 4k^2}\right) + 2k \left(-\frac{16k}{1 + 4k^2}\right) + 4$ $$$$

\cdot \vec{OB} = \frac{12 + 12k^2 - 32k^2 + 4(1 + 4k^2)}{1 + 4k^2} = \frac{16 - 4k^2}{1 + 4k^2}7. **Solve for $k$ using $\vec{OA} \cdot \vec{OB} = 0$**: \frac{16 - 4k^2}{1 + 4k^2} = 0 \implies 16 - 4k^2 = 0 \implies k^2 = 4 \implies k = \pm 2$$ 8. Check discriminant: $k^2 = 4 > \frac{3}{4}$, which satisfies $\Delta > 0$. Hence, $k = \pm 2$.