Center at (2,-5), conjugate axis parallel to the y-axis, slopes at asymptotes numerically one-sixteenth times the length of the latus rectum, and distance between foci is 2sqrt145.
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To determine the equation of the hyperbola, we need the following information:
1. Center: (h, k)
2. Conjugate axis: This axis is parallel to the y-axis and extends in the y-direction.
3. Slopes of the asymptotes: m = ±(b/a)
4. Length of the latus rectum: 2b
5. Distance between the foci: 2c
Given Information:
Center: (2, -5)
Conjugate axis parallel to the y-axis
Slopes of the asymptotes (m): ±(1/16) * latus rectum
Distance between the foci: 2√145
Let's find the missing information step-by-step:
Step 1: Length of the latus rectum (2b)
The latus rectum is given by the formula: L = 2b = 2c * e
where e is the eccentricity of the hyperbola and is given by the formula: e = c/a
From the given information, we know that 2c = 2 * √145, so c = √145
Now, e = c/a = √145/a
Step 2: Eccentricity (e)
The eccentricity is given by the formula: e = c/a
From the previous step, we know e = √145/a
Using the definition of eccentricity, we can write:
e = √(1 + (b^2/a^2))
Step 3: Slopes of the asymptotes (m)
The slopes of the asymptotes are given by m = ±(b/a).
From the given information, we know m = (1/16) * latus rectum = b/a.
Step 4: Determining the value of a
From the definition of eccentricity, we have:
√(1 + (b^2/a^2)) = √145/a
Square both sides:
1 + (b^2/a^2) = 145/a^2
Rearranging the equation:
(b^2/a^2) = (145/a^2) - 1
(b^2/a^2) = (144/a^2)
b^2 = 144, since a^2 cancels out
Taking the square root of both sides:
b = ±12
Step 5: Determining the value of a
Using the slope of the asymptotes:
m = (1/16) * latus rectum = b/a
(1/16) * 2b = b/a
1/8 = 1/a
a = 8
Step 6: Determining the equation of the hyperbola
Given:
Center: (2, -5)
a = 8 (semi-major axis)
b = ±12 (semi-minor axis)
We know the equation of a hyperbola with the center at (h, k) is:
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
Plugging in the values:
(x-2)^2/8^2 - (y+5)^2/12^2 = 1
Therefore, the equation of the hyperbola is:
(x-2)^2/64 - (y+5)^2/144 = 1
To determine the equation of the hyperbola with the given information, we can use the standard form equation for a hyperbola with its center at (h, k) and a conjugate axis parallel to the y-axis:
[(y - k)² / b²] - [(x - h)² / a²] = 1
where:
(h, k) is the center of the hyperbola,
a is the distance between the center and each vertex,
b is the distance between the center and each co-vertex.
1. Center: Given that the center is at (2, -5), we have h = 2 and k = -5.
2. Conjugate axis parallel to the y-axis: This means that the hyperbola opens upwards and downwards. Therefore, the equation involves the variable y.
3. Slopes of asymptotes: It is stated that the slopes of the asymptotes are numerically one-sixteenth times the length of the latus rectum. The latus rectum is the line segment passing through the foci of the hyperbola, and its length is 2b. Therefore, the slopes of the asymptotes are 1/16 * 2b = b/8.
4. Distance between foci: The distance between the foci is given as 2√145.
To find the values of a and b, we need additional information. It seems we are missing some crucial data in order to determine the equation of the hyperbola.