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Find the center,vertices,foci,and asymptotes of the hyperbola.

Find the center,vertices,foci,and asymptotes of the hyperbola.-example-1
User Paulgreg
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2 Answers

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Answer:

The center is (8 , -9)

The vertices are (11 , -9) and (5 , -9)

The foci are (8 , -9 + √58) and (8 , -9 - √58)

The equations of the asymptotes are y = 3/7(x − 8) - 9 , y = -3/7 (x − 8) - 9

Explanation:

- The standard form of the equation of a hyperbola with

center (h , k) and transverse axis parallel to the y-axis is

(y - k)²/a² - (x - h)²/b² = 1

- The length of the transverse axis is 2 a

- The coordinates of the vertices are ( h ± a , k )

- The length of the conjugate axis is 2 b

- The coordinates of the co-vertices are ( h , k ± b )

- The coordinates of the foci are (h , k ± c), where c² = a² + b²

- The equations of the asymptotes are y = ± a/b (x − h) + k

* Now lets solve the problem

∵ (y + 9)²/9 - (x - 8)²/49 = 1

∴ h = 8 and k = -9

∴ a² = 9 ⇒ a = ± 3

∴ b² = 49 ⇒ b = ± 7

∵ c² = a² + b²

∴ c² = 9 + 49 = 58

∴ c = ± √58

∵ The center is (h , k)

∴ The center is (8 , -9)

∵ The coordinates of the vertices are ( h ± a , k )

∴ The vertices are (8 + 3 , -9) and (8 - 3 , -9)

∴ The vertices are (11 , -9) and (5 , -9)

∵ The coordinates of the foci are (h , k ± c)

∴ The foci are (8 , -9 + √58) and (8 , -9 - √58)

∵ The equations of the asymptotes are y = ± a/b (x − h) + k

∴ The equations of the asymptotes are y = 3/7 (x - 8) - 9 and

y = -3/7 (x − 8) - 9

User Select
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Answer:

Center = (-9,8)

Foci = (0,±7.6)

Vertices = (0,±3)

Asymptotes y = 8±(3/7)(x+9)

Explanation:

We need to find the center, vertices, foci and asymptotes of hyperbola:


((y+9)^2)/(9) - ((x-8)^2)/(49)=1

The hyperbola has vertical transverse axis having standard equation:


((y-k)^2)/(a^2) - ((x-h)^2)/(b^2)=1

The center is (h,k), foci (0,±c) , vertices = (0,±a) and

asymptotes = y= k±(a/b)(x-h)

Solving for the given equation by comparing with standard equation:

a^2 = 9 => a = 3

b^2 = 49 => b =7

h= -9

k= 8

c^2 - a^2 = b^2

c^2 = b^2 + a^2

c^2 = 49+9

c^2 = 58

c = 7.6

Now Center(h,k) = (-9,8)

Vertices (0, ±a) = (0,±3) or (0,+3), (0,-3)

Foci (0,±c) = (0, ±7.6) or (0+7.6), (0,-7.6)

Asymptotes = y= k±(a/b)(x-h)

Putting values:

y= 8±(3/7)(x-(-9)

y = 8±(3/7)(x+9)

or y = 8+(3/7)(x+9) and y= 8-(3/7)(x+9)

User Iswanto San
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