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Mathematics 18 Online
OpenStudy (lovelyharmonics):

parabolas

OpenStudy (lovelyharmonics):

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

ganeshie8 (ganeshie8):

ganeshie8 (ganeshie8):

use the chart

OpenStudy (lovelyharmonics):

so the verticies are going to be +-3,0

ganeshie8 (ganeshie8):

Correct ! graphing it may help :)

OpenStudy (lovelyharmonics):

wait but the foci should be +-7 shouldnt it?

OpenStudy (lovelyharmonics):

but my only answers are +-5,0 or +-3,0 :/

ganeshie8 (ganeshie8):

your hyperbola : \(\large \dfrac{x^2}{9} - \dfrac{y^2}{16} = 1\)

ganeshie8 (ganeshie8):

which is same as : your hyperbola : \(\large \dfrac{x^2}{3^2} - \dfrac{y^2}{4^2} = 1\)

ganeshie8 (ganeshie8):

\(\implies a = 3, b = 4\)

ganeshie8 (ganeshie8):

So, vertices = \((\pm a, 0) = (\pm 3, 0) \)

ganeshie8 (ganeshie8):

Foci = \((\pm c, 0) = ?\)

ganeshie8 (ganeshie8):

Find \(c\), by using below relation : \(c^2 = a^2 + b^2\)

OpenStudy (lovelyharmonics):

oh okay that make sense c:

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