The locus of a point that moves such that the difference of its distances from two fixed points called the foci is constant. The constant difference is equal to 2a, which is the length of the transverse axis.

**General Equation:**

Ax^2 - Cy^2 + Dx + Ey + F = 0

**Standard Equation:**

Hyperbola with center at (h, k):

Transverse Axis Horizontal

(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1

Transverse Axis Vertical

(y-k)^2 / a^2 - (x-h)^2 / b^2 = 1

**KEY FORMULAS AND SHORTCUTS:**

** 2a = Length of transverse axis**

** 2b = Length of conjugate axis**

** 2c = distance between the foci**

2b^2 / a **= Length of Latus Rectum**

** d = ** a/e ** = distance from center to directrix**

** e = c/a = eccentricity of hyperbola (greater than 1 for hyperbola)**

c^2 = a^2 + b^2** where c is the distance from center to focus**

From the General Equation Ax^2 - Cy^2 + Dx + Ey + F = 0, we have some shortcuts

**Distance from Focus to Center (c) = ** sqrt ( A + C )

** Center (h, k) = **(-D / 2A, E / 2C)

**Special Note:**

To find the equation of the asymptotes of the hyperbola, we can use the standard equation and equate it to 0 instead of 1. We can then get two equations by using the property of sum and difference. (a^2 - b^2) = (a+b)(a-b)

**Examples and Detailed Solutions:**

1. What is the general equation of a hyperbola with Center at (1, 1), vertex at (1, 5) and conjugate axis 6?

Solution:

We can solve first for a since the center and the vertex is given. We just need to get the distance.

Thus a = 4.

Also, conjugate axis is equal to 2b. Thus 2b = 6 or b = 3.

c^2 = a^2 + b^2

c^2 = 4^2 + 3^2

c = 5

If we graph this, we will notice that the transverse axis is vertical. (Since there was no change on x in vertex and center)

(y-h)^2 / a^2 - (x-h)^2 / b^2 = 1 with (h, k) at (1, 1)

(y-1)^2 / 4^2 - (x-1)^2 / 3^2 = 1

9(y-1)^2 - 16(x-1)^2 = 16(9)

9(y^2 – 2y + 1) - 16(x^2-2x+1)=400

16x^2 - 9y^2 - 32x + 18y + 151 = 0

2**.** Find the equation of the locus of a point which moves so that the difference of its distances from the points (-3, 2) and (7, 2) is 8.

Solution:

Based from the definition, the two fixed points are the foci. Thus, the distance between them is 2c. And the difference is what we called as the transverse axis or 2a.

Distance from (-3, 2) and (7, 2) is 10.

2c = 10

c = 5

2a = 8

a = 4

c^2 = a^2 + b^2

5^2 = 4^2 + b^2

b = 3

To get the Center, get the midpoint of the foci. Midpoint = ( (x1+x2) / 2, (y1+y2) / 2 ) = (-3+7} / 2, (2+2) / 2 )

C is @ (2, 2).

Since the y-value of the foci didn’t change, the transverse axis is horizontal.

(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1 with (h, k) at (2, 2)

(x-2)^2 / 4^2 - (y-2)^2 / 3^2 = 1

9(x-2)^2 - 16(y-2)^2 = 16(9)

9(x^2 – 4x + 4) - 16(y^2-4y+ 4) = 144

9x^2 - 16y^2 - 36x + 64y - 172 = 0

Practice Problems:

1. What is the eccentricity of the hyperbola x^2 - 3y^2-2x+6y + 10 = 0?

2. What is the general equation of the hyperbola with vertices at (-3, 0) and (3, 0) and b = 4?

3. What is the general equation of the hyperbola with Foci at (0, 0) and (0, 10), asymptote x + y = 5?

Answers:

1. sqrt 2

2. 16x^2 - 9y^2 - 144 = 0

3. 2x^2 - 2y^2 + 20y - 25 = 0

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The locus of a point that moves such that the difference of its distances from two fixed points called the foci is constant. The constant difference is equal to 2a, which is the length of the transverse axis.

**General Equation:**

Ax^2 - Cy^2 + Dx + Ey + F = 0

**Standard Equation:**

Hyperbola with center at (h, k):

Transverse Axis Horizontal

(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1

Transverse Axis Vertical

(y-k)^2 / a^2 - (x-h)^2 / b^2 = 1

**KEY FORMULAS AND SHORTCUTS:**

** 2a = Length of transverse axis**

** 2b = Length of conjugate axis**

** 2c = distance between the foci**

2b^2 / a **= Length of Latus Rectum**

** d = ** a/e ** = distance from center to directrix**

** e = c/a = eccentricity of hyperbola (greater than 1 for hyperbola)**

c^2 = a^2 + b^2** where c is the distance from center to focus**

From the General Equation Ax^2 - Cy^2 + Dx + Ey + F = 0, we have some shortcuts

**Distance from Focus to Center (c) = ** sqrt ( A + C )

** Center (h, k) = **(-D / 2A, E / 2C)

**Special Note:**

To find the equation of the asymptotes of the hyperbola, we can use the standard equation and equate it to 0 instead of 1. We can then get two equations by using the property of sum and difference. (a^2 - b^2) = (a+b)(a-b)

**Examples and Detailed Solutions:**

1. What is the general equation of a hyperbola with Center at (1, 1), vertex at (1, 5) and conjugate axis 6?

Solution:

We can solve first for a since the center and the vertex is given. We just need to get the distance.

Thus a = 4.

Also, conjugate axis is equal to 2b. Thus 2b = 6 or b = 3.

c^2 = a^2 + b^2

c^2 = 4^2 + 3^2

c = 5

If we graph this, we will notice that the transverse axis is vertical. (Since there was no change on x in vertex and center)

(y-h)^2 / a^2 - (x-h)^2 / b^2 = 1 with (h, k) at (1, 1)

(y-1)^2 / 4^2 - (x-1)^2 / 3^2 = 1

9(y-1)^2 - 16(x-1)^2 = 16(9)

9(y^2 – 2y + 1) - 16(x^2-2x+1)=400

16x^2 - 9y^2 - 32x + 18y + 151 = 0

2**.** Find the equation of the locus of a point which moves so that the difference of its distances from the points (-3, 2) and (7, 2) is 8.

Solution:

Based from the definition, the two fixed points are the foci. Thus, the distance between them is 2c. And the difference is what we called as the transverse axis or 2a.

Distance from (-3, 2) and (7, 2) is 10.

2c = 10

c = 5

2a = 8

a = 4

c^2 = a^2 + b^2

5^2 = 4^2 + b^2

b = 3

To get the Center, get the midpoint of the foci. Midpoint = ( (x1+x2) / 2, (y1+y2) / 2 ) = (-3+7} / 2, (2+2) / 2 )

C is @ (2, 2).

Since the y-value of the foci didn’t change, the transverse axis is horizontal.

(x-h)^2 / a^2 - (y-k)^2 / b^2 = 1 with (h, k) at (2, 2)

(x-2)^2 / 4^2 - (y-2)^2 / 3^2 = 1

9(x-2)^2 - 16(y-2)^2 = 16(9)

9(x^2 – 4x + 4) - 16(y^2-4y+ 4) = 144

9x^2 - 16y^2 - 36x + 64y - 172 = 0

Practice Problems:

1. What is the eccentricity of the hyperbola x^2 - 3y^2-2x+6y + 10 = 0?

2. What is the general equation of the hyperbola with vertices at (-3, 0) and (3, 0) and b = 4?

3. What is the general equation of the hyperbola with Foci at (0, 0) and (0, 10), asymptote x + y = 5?

Answers:

1. sqrt 2

2. 16x^2 - 9y^2 - 144 = 0

3. 2x^2 - 2y^2 + 20y - 25 = 0

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