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# Some more hyperbola Ebola

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Hyperbola ebola has struck me again! The only cure is an explanation of ths Hyperbola question with steps on how it is done!

$$y^2-9x^2-4y+18x-14=0$$

find the:

center

foci

asymototes

and vertices

graph

Remember the Hyperbola ebola will never be cured without help that has no explained steps, otherwise the brain won't be able to learn the way to fight it off!

Thank you!!!

Apr 7, 2018

#1
+100790
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$$y^2-9x^2-4y+18x-14=0\\ y^2-4y-9x^2+18x=14\\ (y^2-4y)-9(x^2-2x)=14\\ (y^2-4y+4)-9(x^2-2x+1)=14+4-9\\ (y^2-2)^2-9(x-1)^2=9\\ \frac{(y^2-2)^2}{3^2}-\frac{(x-1)^2}{1^2}=1\\$$

Now it is in standard equation form for a hyperbola with a vertical transverse axis.

For the rest see here.

http://www.sparknotes.com/math/precalc/conicsections/section4/

Apr 8, 2018
#2
+91
+3

Ah yes, I am cured!

dom6547  Apr 8, 2018