Question
For
(a) Identify the center,
(b) Identify the vertices.
© Identify the foci.
(d) Write equations for the asymptotes.
(e) Graph the hyperbola.
Give exact answers. Express numbers in simplest form using integers, fractions, or radicals.
Part 1 of 5
(a) The center is (
7 .
).
Part 2 of 5
(b) The vertices are
and
.
Part 3 of 5
© The foci are
and
.
Part:
Part 4 of 5
(d) The asymptotes are
and
.
(a) Identify the center,
(b) Identify the vertices.
© Identify the foci.
(d) Write equations for the asymptotes.
(e) Graph the hyperbola.
Part 1 of 5
(a) The center is (
(b) The vertices are
© The foci are
(d) The asymptotes are
Ask by Haynes Harrington. in the United States
Mar 25,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
Center: (7, 0)
Vertices: (7, 1) and (7, -1)
Foci: (7, √19) and (7, -√19)
Asymptotes: y = 3√2 and y = -3√2
Solution
To identify the center, vertices, foci, and asymptotes of the hyperbola given by the equation
, we need to rewrite the equation in standard form.
The standard form of a hyperbola centered at (h, k) is given by:
Comparing the given equation with the standard form, we can see that the center of the hyperbola is (7, 0).
Now, let’s identify the vertices, foci, and asymptotes:
(a) The center of the hyperbola is (7, 0).
(b) The vertices of the hyperbola are (7, 1) and (7, -1).
© The foci of the hyperbola are (7, √19) and (7, -√19).
(d) The asymptotes of the hyperbola are y = ±√(18/1) = ±√18 = ±3√2.
Therefore, the center is (7, 0), the vertices are (7, 1) and (7, -1), the foci are (7, √19) and (7, -√19), and the asymptotes are y = ±3√2.
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Mind Expander
The center of the hyperbola given by the equation
is located at the point (7, 0). This is derived from the standard form of the hyperbola, where the center
corresponds to the values in the expression. Since there are no
terms subtracted from
, the
coordinate of the center remains 0.
The vertices of the hyperbola are situated at (7, 1) and (7, -1). This is because the distance from the center to the vertices is given by
, which in this case is equal to 1 (derived from the equation). Therefore, we move up and down from the center at (7, 0) by 1 unit to identify the vertices.