Using the information given below write the equation of the hyperbola:

Question

a). The hyperbola has vertices (–2, 9) and (–2, 3) and foci (–2, 13) and (–2, –1).

b). The transverse axis of the hyperbola lies on the line y=–3 and has length 6; the conjugate axis lies on the line x=2 and has length 8.

PLEASE HELP AND THANK YOU IF YOU DO!!

Answer 1

Part a)

Since the vertices (-2, 9) and (-2, 3) have the same x-coordinate, we can state that the hyperbola is vertical. If they had the same y-coordinate, then the hyperbola would be horizontal. So, we can use the equation for a vertical hyperbola which is:

(y - k)²/a² - (x - h)²/b² = 1

We need to find (h, k) which is the center or the midpoint of the vertices. Then we need to find a² and b². To find (h, k) which is the midpoint, we can use the midpoint formula:

((x₁ + x₂)/2, (y₁ + y₂)/2)

((-2 + (-2))/2, (9 + 3)/2)

(-2, 6)

So, (h, k) = (-2, 6)

Next, a is the distance from the center to a vertex so we can just subtract the y-coordinates of the center and a vertex to get that:

a = 9 - 6 = 3

a² = 3² = 9

Next, c is the distance from the center to a focus. We are given that the foci is (–2, 13) and (–2, –1). So, we can subtract the y-coordinates of the focus from the center to find c:

c = 13 - 6 = 7

c² = 49

We found c so that we can now plug it into the hyperbola formula and get b which is:

c² = a² + b²

49 = 9 + b²

b² = 40

Now, we can plug these values into the hyperbola equation:

(y - 6)²/9 - (x - (-2))²/40 = 1

Subtracting a negative becomes addition, so the equation simplifies to:

(y - 6)²/9 - (x + 2)²/40 = 1

Part b)

We are given that The transverse axis of the hyperbola lies on the line y = –3 and has length 6; the conjugate axis lies on the line x = 2 and has length 8.

Lets start by finding (h, k) which is the center or the midpoint of the vertices. Then we need to find a² and b².

To find the center, we know that the transverse axis line lies on y = -3 and conjugate axis line lies on x = 2. The center is where these two lines cross. The line x = 2 and the line y = -3 cross at (2, -3) which would be the center.

Now, the transverse axis tells us the direction in which the hyperbola opens. The transverse axis lies on y = -3, which is a horizontal line. Therefore, this hyperbola opens left and right, so we use the hyperbola horizontal equation:

(x - h)²/a² - (y - k)²/b² = 1

Lets find a which comes from the transverse axis because the transverse axis goes through the vertices. The entire transverse axis has length 6. The center divides it into two equal parts, so:

2a = 6

a = 3

a² = 3² = 9

To find b, we have to know that b comes from the conjugate axis. The entire conjugate axis has length 8. The center divides it into two equal parts, so:

2b = 8

b = 4

b² = 4² = 16

Lets substitute these values into the equation:

(x - 2)²/9 - (y - (-3))²/16 = 1

Subtracting a negative becomes addition, so the equation simplifies to:

(x - 2)²/9 - (y + 3)²/16 = 1

Answer 2

a) Using the vertices, we can find the center of the hyperbola. It is half way between them. It would be (-2,6). We can also use the vertices to get the length of the traverse axis, it would be 9-3 = 6/2 = a=3

Using the foci, we can find the length of the c term and that will allow us to calculate the b² term based on this formula b²=c²-a². 13-(-1) = 14/2 = 7 = c

b²=7²- 3² = 40

A hyperbola will always use subtraction. It will also always =1. The traverse axis is vertical, so the y term will be the first term.

The result:

(y-6)²/9 - (x+2)²/40 = 1

b) By giving the traverse and conjugate axis lines, you are able to figure out the center of the hyperbola. (2,-3) would be the center.

The length of the axis gives you the value for the denominators. They length needs to be divided by 2 and then squared. Denominator for the traverse axis is 6/2 = 3² and the conjugate axis is 8/2 = 4²

A hyperbola will always use subtraction. It will also always =1. The traverse axis is horizontal, so the x term will be the first term.

The result:

(x-2)²/16 - (y+3)²/9 = 1