1) Since the center is at (3, -1) and a vertex is at (5, -1), it opens horizontally.
The general equation will be: (x - h)²/a² - (y - k)²/b² = 1
with center = (h, k), a = length of transverse radius, b = length of conjugate radius
For this problem: h = 3, k = -1, a = 2
The asymptote is 2y = 3x - 11 ---> y = (3/2)x - (11/2) ---> m = 3/2
Since m = 3/2, for every run of 2, it has a rise of 3 ---> Since a = 3, b = 2.
---> (x - 3)²/2² - (y - -1)²/2² = 1
Finish this, and you can choose the answer.
2) The general equaiton will be: (x - h)²/a² - (y - k)²/b² = 1
with center (h, k), a = x-radius, b = y-radius c = focal length
Since foci are at (1, -1) and (1, 5), the center will be at (1, 2) ---> h = 1, k = 2, c = 3
---> (x - 1)²/a² - (y - 2)²/b² = 1
Since the point (4, 2) is on the graph ---> (4 - 1)²/a² - (2 - 2)²/b² = 1,
solve this equation to find the value of a.
To find the value of b, use the equation: b² = a² + c².
(Use that equation if the major axis is vertical; if the major axis is horizontal, use a² = b + c².)
Place those values into the general equation to get your answer.
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