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

Since the distance between the vertices is 16, the denominator of the x-term will be half that, squared (64). Looking at the options which fit that condition (a and c), we wee that only one of those has the correct a in (y - a) to adjust the graph up one unit (answer a). For this particular problem, you don't need to explicitly find the denominator of the y term, which is good, because I don't remember the easy way of doing it.

Answer: 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 placing out (because of the fact the vertices/foci lie on a horizontal line) and the transverse axis is 6 contraptions long. 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) comparable to a million 3) y + 3 = ok(x + 4)² Plug interior the given factor –30 + 3 = ok(–a million + 4)² –27 = 9k ok = –3 y = –3(x + 4)² – 3 you additionally can discover a horizontal parabola assembly the comparable circumstances, so the question is ambiguous

Answer 3

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