Foci of the graph please

Algebra
Tutor: None Selected Time limit: 1 Day

What are the foci of the graph x^2 – 4y^2 = 16?

May 16th, 2015

x^2*-4y^2=16

divide all by 16

x^2/16-4y^2/16=16/16

x^2/16-y^2/4=1

This means our general form for this equation is
x^2/a^2 + y^2/b^2 = 1 

The foci is the following sets of points:
(-c, 0) and (c, 0) 

To find c,
c^2 = a^2 - b^2
The square root of 16 = 4 = a^2.
The square root of 4 = 2 = b^2
Then:
c^2 = 4 - 2
c^2 = 2
Take square root of both sides:
c = +2, -2
Then the following two points for the foci:
(2, 0) and (-2, 0)


May 16th, 2015

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May 16th, 2015
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May 16th, 2015
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