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Find an equation in traditional develop for the hyperbola via 0 votes

vertices at (0, ±2) and also foci at (0, ±7).

You are watching: Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).

hyperbolaaskedJul 11, 2013in PRECALCULUSby homeworkhelpMentor
Please log in or register to include a comment. Please log in or register to answer this question. 2 Answers 0 votes

Given hyperbola

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General equation of a hyperbola

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We have foci and also vertices are on the y-axis, which suggests that we demands the formula for a up and also down hyperbola.

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This suggests that the facility (h, k) must be a the origin, or (0, 0). So, let"s label that...

h = 0k = 0

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We understand that "a" is the distance from your vertex and "c" is the distance from your foci

a = 2 and also c = 7

We have a formula that

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Now fill that

h = 0

k = 0

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Therefore the required equation of hyperbola is

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answeredJul 11, 2013by jouisApprattract Please log in or register to include a comment. 0 votes

The vertices of the hyperbola are (0, 2) and (0, - 2) and its foci are (0, 7) and (0, - 7).

Due to the fact that the x - coordinate isconstantin theverticesandfoci, this is a vertical hyperbola.

The traditional create of vertical hyperbola (y - k)2/a2 - (x - h)2/b2 = 1.

Wbelow, "b " is the number in the denominator of the positive term, If the x - term is negative, then the hyperbola is vertical.a = semi - transverse axis , b = semi - conjugate axis .Center: (h, k )Vertices: (h , k + a ) and also (h, k - a).Foci: (h , k +c) and (h, k - c).

So, the x coodinate of the center of hyperbola is 0.

vertices : (0, 2) and also (0,- 2)

k + a = 2 ----> (1)

k - a = - 2 ---> (2)

Add the equations (1) & (2).

2k = 0

k = 0

So,the y coordinate of center is 0.

Substitute the k worth in (1),

0 + a = 2

a = 2.

foci : (0, 7) and (0, - 7)

k + c = 7

0 + c = 7

c = 7

c2 = a2 + b2

(7)2 = (2)2 + b2

49 - 4 = b2

b = √45

Substitute the (h , k), a, and b in traditional form of hyperbola equation .

(y - 0)2/22 - (x - 0)2/(√45)2 = 1

(y - 0)2/4 - (x - 0)2/45 = 1.

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Because of this, the standard create of hyperbola is (y - 0)2/4 - (x - 0)2/45 = 1.

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