Question

please help
What are the vertices, foci and asymptotes of the hyperbola with equation 16x^2-4y^2=64

Answer 1

V (2,0) & (-2,0)
Co-V (0,4) & (0,-4)
foci (3,0) & (-3,0)
asymptotes y=-1/2x and y = 1/2x

Answer 2

how did you find them?

Answer 3

oh your equation is x^2/4 - y^2/16 = 1 when you divide both sides with 64.then you find the a and a is the denominator before the subtraction sign. so a is 2, b is the denom. after the subtraction sign so its 4. then you use a^2 + b^2 = c^2 to find c which is +/_ 3

Answer 4

then since a is under the x it means its a horizontal hyperbola. center is (0,0) so for the V add +/- 2 to the x leaving you with V(+/-2,0) then add the 4 to the y in (0,0) ->
(0, +/-3) as your co vertices. also since your focus has to be on the same line as your V you add your +/-3 to the X of (0,0) leaving you with (+/-3,0) . then you can find your asymptote by using the formula y=+/- a/bx (for horizontal only)

Answer 5

was that helpful

Answer 6

@NataliaK Yes thank you!! :)

Answer 7

:) glad I could help