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hyperbola conics question (1 Viewer)
- Thread starter kr73114
- Start date
ninetypercent
ninety ninety ninety
1) foci are on x axis.
ae = 10
b/a = 4/3
b^2 = a^2(e^2-1)
b^2/a^2 = e^2 - 1
16/9 + 1 = e^2
25/9 = e^2
e = 5/3 (e>0)
a = 10/e
= 6
b/6 = 4/3
b = 4/3 x 6
= 8
equation is x^2/36 - y^2/64 = 1
ae = 10
b/a = 4/3
b^2 = a^2(e^2-1)
b^2/a^2 = e^2 - 1
16/9 + 1 = e^2
25/9 = e^2
e = 5/3 (e>0)
a = 10/e
= 6
b/6 = 4/3
b = 4/3 x 6
= 8
equation is x^2/36 - y^2/64 = 1
ninetypercent
ninety ninety ninety
haven't touched this stuff in agessssss. I'm starting to forget. correct me if i am wrong LOL
ninetypercent
ninety ninety ninety
2) find the equation of a hyperbola with eccentricity 5/4, focus (-10,0) and
directrix x=-32/5
ae = 10
a x 5/4 = 10
a = 10 x 4/5
= 8
a^2 = b^2(e^2-1)
64 = b^2(25/16-1)
64 = b^2 (9/16)
b^2 = 64 x 16/9
b = 32/3 (i think) didn't use calculator
and then plug into equation
directrix x=-32/5
ae = 10
a x 5/4 = 10
a = 10 x 4/5
= 8
a^2 = b^2(e^2-1)
64 = b^2(25/16-1)
64 = b^2 (9/16)
b^2 = 64 x 16/9
b = 32/3 (i think) didn't use calculator
and then plug into equation
thanks