SOLUTION: Vertices: (15, 1), (-1, 1); Endpoints of Conjugate Axis: (7, 7), (7, -5)

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Question 1186254: Vertices: (15, 1), (-1, 1); Endpoints of Conjugate Axis: (7, 7),
(7, -5)

Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

Vertices: (, ), (, );
Endpoints of Conjugate Axis: (, ), (, )
-> the center is at midpoint (, )=> C (,)
the length of Conjugate Axis is =>

equation of hyperbola
........plug in known


........use vertices (, ) to calculate




so, your equation is:




Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.
Vertices: (15, 1), (-1, 1); Endpoints of Conjugate Axis: (7, 7), (7, -5).

~~~~~~~~~~~~~~~~~~~

Since the conjugate axis is mentioned, it is about hyperbola.


Its center is the point  (7,1).


It has horizontal major axis y = 1, parallel to x-axis.

The distance between vertices is  15 - (-1) = 16;  hence, the major semi-axis length  "a"  is  16/2 = 8  units.


It has vertical conjugate axis  x = 7, parallel to y-axis.

The length of the conjugate axis is 7 - (-5) = 12;  hence, the minor semi-axis length  "b"  is  12/2 = 6  units.



Combining all this info, the standard equation of the hyperbola is


     -  = 1.

Solved.



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