What is the equation of the hyperbola with foci (0,+-10) and passing through (2,3)?

Question

dont try being smart mister..ive got a board exam tomorrow!

how do i find the eccentricity!

Freakin jackasses...dont knw d answer just shut up okay!

Answer 1

Find the equation of the hyperbolas (they come in pairs) with foci (0,±10) and passing thru the point (2,3).

The midpoint between the two foci is the center (h,k).

(h,k) = ((0+0)/2,(10+(-10))/2) = (0,0)

The foci run vertically so the hyperbolas open up and down. The equation of the hyperbolas is therefore of the form:

y²/a² - x²/b² = 1

The distance between the two foci is 2c.

2c = 10 - (-10) = 10 + 10 = 20
c = 10

The absolute value of the difference of the distance from the foci to a point on one of the hyperbolas is a constant "2a".

2a = √[(2 - 0)² + (-10 - 3)²] - √[(2 - 0)² + (10 - 3)²]
2a = √(4 + 169) - √(4 + 49)
2a = √173 - √53
a = (√173 - √53)/2

a² = (173 - 2(√173)(√53) + 53)/4
a² = (226 - 2√9169)/4
a² = (113 - √9169)/2 ≈ 8.6225523

b² = c² - a²
b² = 10² - (113 - √9169)/2
b² = 100 - (113 - √9169)/2
b² = (87 + √9169)/2 ≈ 91.377448

The equation of the hyperbolas is therefore:

y²/a² - x²/b² = 1
y² / [(113 - √9169)/2] - x² / [(87 + √9169)/2] = 1
_____________________

Eccentricity is e = c/a. For hyperbolas e > 1.
e = c/a = 10 / [(√173 - √53)/2]
e = 20 / (√173 - √53) ≈ 3.4055094
____________________________

As an aside, just because some of the answers before me were nasty to you is not reason for you to be nasty back. I almost didn't answer the question for that reason. Good luck on your boards.