The Hyperbola

Definition of a Hyperbola

A hyperbola is the locus of a point that moves in a plane such that the absolute difference of its distances from two fixed points (the foci) is constant. This constant difference is equal to the length of the transverse axis (2a).

$$|PF_1 - PF_2| = 2a$$

Alternatively, it can be defined as a conic section with an eccentricity e such that e > 1. This means any point on the hyperbola is farther from the directrix than it is from the focus.

Standard Equation of a Hyperbola

The standard equation of a hyperbola centered at the origin is characterized by a minus sign between the terms.

• Horizontal Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

  • Vertices: $(\pm a, 0)$
  • Foci: $(\pm c, 0)$, where $c = ae$.
  • Transverse Axis: The axis connecting the vertices, length $2a$ (along the x-axis).
  • Conjugate Axis: Length $2b$ (along the y-axis).

• Vertical Hyperbola: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

  • Vertices: $(0, \pm a)$
  • Foci: $(0, \pm c)$, where $c = ae$.
  • Transverse Axis: Length $2a$ (along the y-axis).
  • Conjugate Axis: Length $2b$ (along the x-axis).

Key Terminology and Formulas

• Relationship between a, b, and c: Unlike the ellipse, the relationship for a hyperbola is:

$$c^2 = a^2 + b^2$$ The eccentricity relationship is: $$b^2 = a^2(e^2-1)$$

• Directrices: The equations of the two directrices are $x = \pm \frac{a}{e}$.
• Latus Rectum: The focal chord perpendicular to the transverse axis. Its length is $\frac{2b^2}{a}$.
• Asymptotes: These are crucial lines that the hyperbola's branches approach at infinity. They form a 'guide' for the shape of the curve. Their equations are $y = \pm \frac{b}{a}x$ for a horizontal hyperbola.

Rectangular Hyperbola

A hyperbola is rectangular (or equilateral) if its asymptotes are perpendicular.

• Condition: This happens when $a = b$. Its eccentricity is always $e=\sqrt{2}$.
• Equation: The standard form is $x^2-y^2=a^2$. When the coordinate axes are rotated by 45° to align with the asymptotes, the equation becomes $xy = c^2$.

Example 1: Finding Parameters from the Equation

Question: For the hyperbola $9x^2 - 16y^2 = 144$, find the coordinates of the foci and vertices, the eccentricity, and the length of the latus rectum.

Solution: Step 1: Divide by 144 to get the standard form: $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Step 2: Identify parameters. This is a horizontal hyperbola with $a^2=16 \implies a=4$ and $b^2=9 \implies b=3$.

• Vertices: $(\pm a, 0) = \mathbf{(\pm 4, 0)}$.
• Find c: $c^2 = a^2+b^2 = 16+9=25 \implies c=5$.
• Foci: $(\pm c, 0) = \mathbf{(\pm 5, 0)}$.
• Eccentricity: $e=c/a = \mathbf{5/4}$.
• Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \mathbf{9/2}$.

Example 2: Finding the Equation from Parameters

Question: Find the equation of the hyperbola whose foci are $(\pm 5, 0)$ and the length of the transverse axis is 8.

Solution: Foci are on the x-axis, so it's a horizontal hyperbola. From the foci $(\pm c, 0)$, we get $\mathbf{c=5}$. The length of the transverse axis is $2a=8 \implies \mathbf{a=4}$.

Now find $b^2$ using $c^2=a^2+b^2$: $5^2 = 4^2 + b^2 \implies 25 = 16 + b^2 \implies b^2=9$.

The equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, which is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Example 3: Hyperbola with a Shifted Center

Question: Find the center and foci for the hyperbola $9x^2 - 4y^2 - 18x - 16y - 43 = 0$.

Solution: Step 1: Group terms and complete the square. $9(x^2 - 2x) - 4(y^2 + 4y) = 43$ $9(x^2 - 2x + 1) - 4(y^2 + 4y + 4) = 43 + 9(1) - 4(4)$ $9(x-1)^2 - 4(y+2)^2 = 36$

Step 2: Divide by 36: $\frac{(x-1)^2}{4} - \frac{(y+2)^2}{9} = 1$.

• Center: $(h,k) = \mathbf{(1, -2)}$.
• Parameters: $a^2=4, b^2=9$.
• Foci distance: $c^2 = a^2 + b^2 = 4 + 9 = 13 \implies c=\sqrt{13}$.
• Foci: Horizontal hyperbola, so foci are at $(h \pm c, k)$, which are $(1 \pm \sqrt{13}, -2)$.

Example 4: Using Asymptotes

Question: Find the equation of the hyperbola whose asymptotes are $y = \pm 2x$ and which passes through the point (3, 8).

Solution: The asymptotes are $y=\pm\frac{b}{a}x$, so we have $\frac{b}{a}=2 \implies b=2a$. The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1$.

Substitute $b=2a$: $\frac{x^2}{a^2} - \frac{y^2}{4a^2} = \pm 1$.

The point (3,8) lies on the curve: $\frac{3^2}{a^2} - \frac{8^2}{4a^2} = \pm 1 \implies \frac{9}{a^2} - \frac{64}{4a^2} = \pm 1 \implies \frac{9}{a^2} - \frac{16}{a^2} = \pm 1$.

$\frac{-7}{a^2} = \pm 1$. Since $a^2$ must be positive, we take the negative sign: $\frac{-7}{a^2} = -1 \implies a^2=7$.

Then $b^2 = (2a)^2 = 4a^2 = 4(7)=28$.

The original equation was $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ (a vertical hyperbola).

Equation: $\frac{y^2}{28} - \frac{x^2}{7} = 1$.