Question
Asked Feb 26, 2020
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I am needing assistance with the attached equation. 

Find the center, vertices and foci of the hyperbola.
x2
25
36
a. Center: (0, 0)
(0, 6)
(0, /61)
Vertices:
Foci:
b. Center: (0, 0)
(t6, 0)
Vertices:
Foci: (+v61, 0)
c. Center: (0, 0)
Vertices: (+6, 0)
(61, 0)
Foci:
d. Center: (0, 0)
(6, 0)
Vertices:
Foci: (+V61, 0)
e. Center: (0, 0)
(0, +5)
Vertices:
(0, ±/61)
Foci:
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Find the center, vertices and foci of the hyperbola. x2 25 36 a. Center: (0, 0) (0, 6) (0, /61) Vertices: Foci: b. Center: (0, 0) (t6, 0) Vertices: Foci: (+v61, 0) c. Center: (0, 0) Vertices: (+6, 0) (61, 0) Foci: d. Center: (0, 0) (6, 0) Vertices: Foci: (+V61, 0) e. Center: (0, 0) (0, +5) Vertices: (0, ±/61) Foci:

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Expert Answer

Step 1

Given:

Trigonometry homework question answer, step 1, image 1

General equation of hyperbola centred at (h,k) and vertical transverse axis is given as:

Trigonometry homework question answer, step 1, image 2

Step 2

Rewriting and comparing the given equation, we get:

Trigonometry homework question answer, step 2, image 1

Vertices of the hyperbola are given as:

Trigonometry homework question answer, step 2, image 2

Step 3

Now,

Trigonometry homework question answer, step 3, image 1

Hence focus of hyperbola is given as:

Trigonometry homework question answer, step 3, image 2

...

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