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Give the foci of x^2/9+y^2/4=1 (sq.rt of____,____),(-sq rt. of___,____)
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the foci is obtained using the value of c in c^2 = a^2 - b^2, where a=3, b=2. once you find the value of c, move c units up and down from the center to get your foci.
can you give the value of c?
c= sq. root of 13
subtract.. don't add...
c^2 = a^2 - b^2 = 3^2 - 2^2 = 9 - 4 = 5 so c = sqrt5
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plus/minus sqrt5
also... sorry 'bout the mistake up there... the center is (0, 0) and you go left/right of the center, NOT up/down of the center.
okay thanks, is it the same for a hyperbola? Do I add or subtract?
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