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zlopas [31]
1 year ago
6

This hyperbola is centered at theorigin. Find its equation.Foci: (-2,0) and (2,0)Vertices: (-1,0) and (1,0)

Mathematics
1 answer:
beks73 [17]1 year ago
5 0

SOLUTION

From the question, the center of the hyperbola is

\begin{gathered} (h,k),\text{ which is } \\ (0,0) \end{gathered}

a is the distance between the center to vertex, which is -1 or 1, and

c is the distance between the center to foci, which is -2 or 2.

b is given as

\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}

But equation of a hyperbola is given as

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

Substituting the values of a, b, h and k, we have

\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}

Hence the answer is

\frac{x^2}{1}-\frac{y^2}{3}=1

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Answer:

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Step-by-step explanation:

Hi there!

Let x be equal to the larger integer.

Let y be equal to the smaller integer.

<u>1) Construct equations</u>

  1. x+y=33 (The sum of two integers is 33)
  2. x=6+2y (The larger is 6 more than twice the smaller)

<u>2) Solve for one of the integer</u>

Isolate x in the first equation

x+y=33\\x=33-y

Plug the first equation into the second

33-y=6+2y

Combine like terms

33-6=2y+y\\27=3y\\9=y

Therefore, the smaller integer is 9.

<u />

<u>3) Solve for the other integer</u>

x+y=33

Plug in y (9)

x+9=33\\x=33-9\\x=24

Therefore, the larger integer is 24.

I hope this helps!

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3 years ago
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Use the Distance Formula.

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