write an equation for the hyperbola with vertices (-10,1) and (6,1) and foci (-12,1) and (8,1)?

Question

a. (x+2)^2 /64 - (y-1)^2 /36=1
b. (x-2)^2 /36 - (y+1)^2 /64=1
c. (x-2)^2 /64 - (y+1)^2 /36=1
d. (x+2)^2 /36 - (y-1)^2 /64=1

Answer 1

The center would be the point midway between the two vertices:

((-10 + 6)/2, 1) = (-2, 1)

So that immediately eliminates choices b and c.

Then the "semi-major axis", or a, is 1/2 the distance between the two vertices: (6 - (-10))/2 = 8

So the 8² = 64 should be associated with the x term, since it's an "east-west" hyperbola.

Therefore, the answer is:

(x + 2)²/64 - (y - 1)²/36 = 1

Choice a.

Answer 2

a million) the middle is 0.5-way between the vertices (or foci) so is (2, 2) The hyperbola is east-west starting up (because the vertices/foci lie on a horizontal line) and the transverse axis is 6 contraptions lengthy. So we've (x – 2)²/36 – (y – 2)²/b² = a million The foci are at a squared distance c² = a² + b² = one hundred from the middle, so b² = one hundred – 36 = sixty 4 The equation is then (x – 2)²/36 – (y – 2)²/sixty 4 = a million 2) resembling a million 3) y + 3 = ok(x + 4)² Plug contained in the given element –30 + 3 = ok(–a million + 4)² –27 = 9k ok = –3 y = –3(x + 4)² – 3 you may also hit upon a horizontal parabola assembly a similar situations, so the question is ambiguous