Answer:
- (x -1)²/64 -(y -4)²/80 = 1
- foci (-11, 4) and (13, 4)
Step-by-step explanation:
The standard form of a hyperbola whose transverse axis is horizontal is ...
(x -h)²/a² -(y -k)²/b² = 1 . . . . center (h, k); transverse axis 2a; conjugate axis 2b
The distance between the two vertices is given as (9 -(-7)) = 16 = 2a, so the value of a is 8 and we have the partial equation ...
(x -1)²/8² -(y -4)²/b² = 1
Filling in the given point value gives us an equation for b².
(-11 -1)²/8² -(-6 -4)²/b² = 1
9/4 -100/b² = 1
5/4 = 100/b²
b² = 80
So, the equation is ...
(x -1)²/64 +(y -4)²/80 = 1
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The distance between the foci is 2c, where c² = a² +b².
c² = 64 +80 = 144 = 12²
c = 12, so the foci are at ...
(x, y) = (1 ±12, 4) . . . . . . . 12 units either side of the center
The foci are (-11, 4) and (13, 4).
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Additional comment
The steps are ...
- use the given information to fill in as much of the equation as you can.
- use any given points or additional information to find the unknown values in the equation
- determine any additional information asked for (foci, asymptotes, ...)
The relevant relations are ...
2a = length of transverse axis (distance between vertices)
2b = length of conjugate axis (distance between co-vertices)
±b/a = slope of asymptotes, which are lines through the center
2c = distance between foci, where c² = a² +b²
The equation is (x -h)²/a² -(y -k)²/b² = 1 for a hyperbola that opens horizontally with center (h, k). Swapping variables x and y will give a hyperbola that opens vertically. If the figure opens vertically, the asymptotes will have slope ±a/b.
The distances from a point to the foci have a constant difference of 2a.
