Question

find the foci

Answer 1

First transform into standard for by dividing by 36 (you want the right side to equal +1)

Answer 2

Hyperbolae with the form below have center (0, 0) and horizontal transverse axis (the hyperbola opens "left/right"
\[\frac{ x^2 }{ a^2 }+\frac{ y^2 }{ b^2 }=1\]

Answer 3

ok so i solve it this is what I got
\[\frac{ x ^{2} }{ 4 } - \frac{ y^{2} }{ 9 } = 1\]

Answer 4

sorry +

Answer 5

Right.

Answer 6

no I was right the first time its -ve , sorry

Answer 7

The vertices (not asked for in the problem) are a units left and right from the center.

Answer 8

ok

Answer 9

The distance of the foci from the center, along the transverse axis, is taken to be c units, where c is in a Pythagorean relationship with a and b:
\[c^2=a^2+b^2\]

Answer 10

so the foci will be (c+_,0) right

Answer 11

exactly

Answer 12

ok so i have a^2=4 and b^2=9 so c^2= 13

Answer 13

correct, and c = sqrt{13}

Answer 14

so (sqrt13,0) and (-sqrt13,0)
correct ?

Answer 15

A

Answer 16

correct

Answer 17

do you know how to tell the orientation of the hyperbola?

Answer 18

i think yes if the -ve sign is in front of the x so it is vertical
if it is infront of the y then it is horizontal

Answer 19

am i correct ?

Answer 20

correct: in standard form, negative in front of the x-term then transverse axis is vertical; negative in front of the y-term then transverse axis is horizontal (this case)

Answer 21

thank you so much for your help and for the tips @gryphon really appreciate it:)

Answer 22

my pleasure!