6.6 Q-2

Question Statement

Find the center, foci, eccentricity, vertices, and equations of the directrices for the given hyperbolas.

---

Background and Explanation

Hyperbolas are conic sections defined as the locus of points where the difference of distances to two fixed points (foci) is constant. In standard form:

• Horizontal hyperbola:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

• Vertical hyperbola:

$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$

Key parameters:

• Vertices: $(\pm a, 0)$ for horizontal, $(0, \pm a)$ for vertical hyperbolas.
• Foci: Points $(\pm c, 0)$ or $(0, \pm c)$, where $c^2 = a^2 + b^2$.
• Directrices: Lines given by $x = \pm \frac{a}{e}$ or $y = \pm \frac{a}{e}$, where $e = \frac{c}{a}$ is the eccentricity.

---

Solution

i. Given: $x^2 - y^2 = 9$

Steps:

1. Rewrite in standard form:

$$\frac{x^2}{9} - \frac{y^2}{9} = 1$$

Here, $a^2 = 9$, $b^2 = 9$, so $c^2 = a^2 + b^2 = 18$, and $c = \sqrt{18} = 3\sqrt{2}$.

2. Identify:
   • Center: $(0, 0)$
   • Vertices: $(\pm 3, 0)$
   • Foci: $(\pm 3\sqrt{2}, 0)$
   • Directrices: $x = \pm \frac{3}{\sqrt{2}}$

---

ii. Given: $\frac{x^2}{4} - \frac{y^2}{9} = 1$

Steps:

1. Identify parameters:

   • $a^2 = 4$, $b^2 = 9$, $c^2 = a^2 + b^2 = 13$, $c = \sqrt{13}$.

2. Identify:

   • Center: $(0, 0)$
   • Vertices: $(\pm 2, 0)$
   • Foci: $(\pm \sqrt{13}, 0)$
   • Directrices: $x = \pm \frac{2}{\sqrt{13}}$

---

iii. Given: $\frac{y^2}{16} - \frac{x^2}{9} = 1$

Steps:

1. Identify parameters:

   • $a^2 = 16$, $b^2 = 9$, $c^2 = a^2 + b^2 = 25$, $c = 5$.

2. Identify:

   • Center: $(0, 0)$
   • Vertices: $(0, \pm 4)$
   • Foci: $(0, \pm 5)$
   • Directrices: $y = \pm \frac{4}{5}$

---

iv. Given: $\frac{y^2}{4} - \frac{x^2}{1} = 1$

Steps:

1. Identify parameters:

   • $a^2 = 4$, $b^2 = 1$, $c^2 = a^2 + b^2 = 5$, $c = \sqrt{5}$.

2. Identify:

   • Center: $(0, 0)$
   • Vertices: $(0, \pm 2)$
   • Foci: $(0, \pm \sqrt{5})$
   • Directrices: $y = \pm \frac{2}{\sqrt{5}}$

---

v. Given: $\frac{(x-1)^2}{2} - \frac{(y-1)^2}{9} = 1$

Steps:

1. Shift center to $(1, 1)$:
   • Substitute $X = x-1$, $Y = y-1$:

$$\frac{X^2}{2} - \frac{Y^2}{9} = 1$$

• Here, $a^2 = 2$, $b^2 = 9$, $c^2 = a^2 + b^2 = 11$, $c = \sqrt{11}$.

2. Identify:
   • Center: $(1, 1)$
   • Vertices: $(1 \pm \sqrt{2}, 1)$
   • Foci: $(1 \pm \sqrt{11}, 1)$
   • Directrices: $x = 1 \pm \frac{\sqrt{2}}{\sqrt{11}}$

---

vi. Given: $\frac{(y+2)^2}{9} - \frac{(x-2)^2}{16} = 1$

Steps:

1. Shift center to $(2, -2)$:
   • Substitute $X = x-2$, $Y = y+2$:

$$\frac{Y^2}{9} - \frac{X^2}{16} = 1$$

• Here, $a^2 = 9$, $b^2 = 16$, $c^2 = a^2 + b^2 = 25$, $c = 5$.

2. Identify:
   • Center: $(2, -2)$
   • Vertices: $(2, -2 \pm 3)$
   • Foci: $(2, -2 \pm 5)$
   • Directrices: $y = -2 \pm \frac{3}{5}$

---

vii. Given: $9x^2 - 12x - y^2 - 2y + 2 = 0$

Steps:

1. Rewrite in standard form:
   • Complete the square:

$$9(x^2 - \frac{4}{3}x) - (y^2 + 2y) = -2$$ $$9(x - \frac{2}{3})^2 - (y + 1)^2 = 1$$

• Divide by 9:

$$\frac{(x - \frac{2}{3})^2}{\frac{1}{9}} - \frac{(y + 1)^2}{1} = 1$$

2. Identify:
   • Center: $(\frac{2}{3}, -1)$
   • Vertices: $(\frac{5}{3}, -1), (\frac{-1}{3}, -1)$
   • Foci: $(\frac{2}{3} \pm \sqrt{10}, -1)$
   • Directrices: $x = \frac{2}{3} \pm \frac{1}{\sqrt{10}}$

---

Key Formulas or Methods Used

1. Relation of parameters: $c^2 = a^2 + b^2$
2. Equations of hyperbolas:
   • Horizontal: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
   • Vertical: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$
3. Directrices: $x = \pm \frac{a}{e}$ or $y = \pm \frac{a}{e}$

---

Summary of Steps

1. Rewrite the equation in standard form.
2. Identify key parameters: $a^2$, $b^2$, $c^2 = a^2 + b^2$.
3. Determine:
   • Center
   • Vertices
   • Foci
   • Directrices
4. Verify orientation (horizontal or vertical) and construct the equation.

---