6.6 Q-1

Question Statement

Find the equations of the hyperbolas for the given conditions. Include all necessary steps, calculations, and sketches for clarity.

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Background and Explanation

A hyperbola is a conic section defined as the locus of points where the absolute difference of distances to two fixed points (foci) is constant. The standard form of a hyperbola depends on the orientation of its transverse axis:

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

The key parameters are:

• $a$: Distance from the center to each vertex.
• $b$: Distance related to the conjugate axis.
• $c$: Distance from the center to each focus, related by $c^2 = a^2 + b^2$.

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Solution

i. Given: Center $(0, 0)$, Focus $(6, 0)$, Vertex $(4, 0)$

Steps:

1. Identify parameters:
   • $a = 4$, $c = 6$.
   • Using $c^2 = a^2 + b^2$, calculate $b^2$:

$$b^2 = c^2 - a^2 = 36 - 16 = 20$$

2. Since the transverse axis is horizontal, the equation of the hyperbola is:

$$\frac{x^2}{16} - \frac{y^2}{20} = 1$$

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ii. Given: Foci $(\pm 5, 0)$, Vertex $(3, 0)$

Steps:

1. Identify parameters:
   • $a = 3$, $c = 5$.
   • Calculate $b^2$:

$$b^2 = c^2 - a^2 = 25 - 9 = 16$$

2. Transverse axis is horizontal; the equation of the hyperbola is:

$$\frac{x^2}{9} - \frac{y^2}{16} = 1$$

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iii. Given: Foci $(2+5\sqrt{2}, 7)$ and $(2-5\sqrt{2}, -7)$, $2a = 10$

Steps:

1. Identify parameters:
   • $a = 5$, $c = 5\sqrt{2}$.
   • Calculate $b^2$:

$$b^2 = c^2 - a^2 = (5\sqrt{2})^2 - 25 = 50 - 25 = 25$$

2. Since the transverse axis is horizontal, the equation is:

$$\frac{(x-2)^2}{25} - \frac{(y+7)^2}{25} = 1$$

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iv. Given: Foci $(0, \pm 9)$, Directrices $y = \pm 4$

Steps:

1. Identify parameters:
   • $c = 9$, $\frac{a}{e} = 4$, so $e = \frac{c}{a} = \frac{9}{4}$ and $a = 4$.
   • Calculate $b^2$:

$$b^2 = c^2 - a^2 = 81 - 16 = 65$$

2. Since the transverse axis is vertical, the equation is:

$$\frac{y^2}{36} - \frac{x^2}{45} = 1$$

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v. Given: Center $(2, 2)$, Horizontal transverse axis of length $6$, Eccentricity $e = 2$

Steps:

1. Identify parameters:
   • $2a = 6 \Rightarrow a = 3$.
   • $c = ae = 6$.
   • Calculate $b^2$:

$$b^2 = c^2 - a^2 = 36 - 9 = 27$$

2. The equation of the hyperbola is:

$$\frac{(x-2)^2}{9} - \frac{(y-2)^2}{27} = 1$$

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vi. Given: Vertices $(2, \pm 3)$, Point on curve $(0, 5)$

Steps:

1. From the vertices:
   • $2a = 6 \Rightarrow a = 3$, $a^2 = 9$.
2. Substitute $(x, y) = (0, 5)$ into the general form:

$$\frac{5^2}{9} - \frac{4}{b^2} = 1$$

Solve for $b^2$:

$$\frac{25}{9} - \frac{4}{b^2} = 1 \Rightarrow \frac{4}{b^2} = \frac{16}{9} \Rightarrow b^2 = \frac{9}{4}$$

3. The equation is:

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

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vii. Given: Vertices $(5, -2)$ and $(5, 4)$, Focus $(5, -3)$

Steps:

1. Identify parameters:
   • Center $(5, 1)$ (midpoint of vertices), $a = 3$, $a^2 = 9$.
   • $c = 4$, $c^2 = 16$.
   • Calculate $b^2$:

$$b^2 = c^2 - a^2 = 16 - 9 = 7$$

2. The equation of the hyperbola is:

$$\frac{(y-1)^2}{9} - \frac{(x-5)^2}{7} = 1$$

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Key Formulas or Methods Used

1. $c^2 = a^2 + b^2$: Relates the distances $a$, $b$, and $c$ in a hyperbola.
2. Horizontal hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
3. Vertical hyperbola: $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.
4. Center: Midpoint of the vertices.

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Summary of Steps

1. Identify $a$, $b$, and $c$ based on the given data.
2. Calculate $b^2$ using $c^2 = a^2 + b^2$.
3. Determine the orientation (horizontal or vertical) of the hyperbola.
4. Write the equation in standard form.
5. Verify with any additional data, such as a point on the curve.

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