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Mathematics 16 Online
OpenStudy (egbeach):

Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1

rishavraj (rishavraj):

equatio of hyperbola is \[\frac{ x^2 }{ 3^2 } - \frac{ y^2 }{ 4^2 } = 1\]

rishavraj (rishavraj):

\[F_1 = (-c , 0) ~~~~and~~~~F_2 = (c , 0)\]

OpenStudy (egbeach):

\[\frac{ x ^{2} }{ 9 }- \frac{ y ^{2} }{ 16 } =1\]

OpenStudy (egbeach):

thats the equation

rishavraj (rishavraj):

9 = 3^2 and 16 = 4^2

rishavraj (rishavraj):

see standard equation of hyperbola is \[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1\]

OpenStudy (egbeach):

ohhh. sorry

rishavraj (rishavraj):

not a big deal....... @egbeach so now \[c^2 = a^2 + b^2 \]

OpenStudy (egbeach):

c^2= 25 so c=5

rishavraj (rishavraj):

so now u can get focii??/

OpenStudy (egbeach):

yeah! thank you

rishavraj (rishavraj):

vertices???

OpenStudy (egbeach):

(+3,0) and (-3,0) right?

rishavraj (rishavraj):

yup :))))

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