Write the equation of a hyperbola with vertices at (1,1) and (9, 1) and foci at (0, 1) and (10, 1).

Question

anonymous · Answer

ok here we go
what is the center? it is half way between $(1,1)$ and $(9,1)$

anonymous · Answer

The length of the transverse axis is the distance between the vertices: 2a = 8, a = 4.

 Since the vertices are on a horizontal line, y = 1, the transverse axis is horizontal, and the hyperbola is of the form:

 (x-g)²/4² - (y-h)²/b² = 1

 The center is halfway between the vertices:
 (g, h) = (10/2, 2/2) = (5, 1)

 (x-5)²/4² - (y-1)²/b² = 1

 The distance between the foci = 2c = 10 => c = 5
 c² = a² + b²
25 = 16 + b² 
b = 3

 The equation of the hyperbola, in "conic" form is:

 (x-5)²/4² - (y-1)/3² = 1

 The equation of the hyperbola, in standard form is:

 9x² - 16y² - 90x + 32y + 65 = 0

anonymous · Answer

you got this, or you want to do it step by step? it is not hard

anonymous · Answer

I think I have it (: thanks

anonymous · Answer

yw