SOLUTION: To two decimal places, what is the value of the eccentricity of the conic with equation : {{{ax^2+by^2=c}}} . Where a= 4,b= -16 and c= 12? I have tried dividing the equation by

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: To two decimal places, what is the value of the eccentricity of the conic with equation : {{{ax^2+by^2=c}}} . Where a= 4,b= -16 and c= 12? I have tried dividing the equation by      Log On


   

Question 959216: To two decimal places, what is the value of the eccentricity of the conic with equation : ax%5E2%2Bby%5E2=c . Where a= 4,b= -16 and c= 12?
I have tried dividing the equation by 12 which gives me x%5E2%2F3-16y%5E2%2F12=1
which makes me stuck here because i can't make it into x%5E2%2Fa%5E2-y%5E2%2Fb%5E2+=1 a hyperbola equation.
please assist.
thank you
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2F3-16y%5E2%2F12=1
x%5E2%2F%28+sqrt%283%29+%29%5E2-16y%5E2%2F12+=+1
Divide top and bottom of the 2nd term by +16+
x%5E2%2F%28+sqrt%283%29+%29%5E2-y%5E2%2F%28%2812%2F16%29%29+=+1
x%5E2%2F%28+sqrt%283%29+%29%5E2-y%5E2%2F%28%283%2F4%29%29+=+1
x%5E2%2F%28+sqrt%283%29+%29%5E2-y%5E2%2F%28%28sqrt%283%2F4%29%29%5E2%29+=+1
+a+=+sqrt%283%29+
+b+=+sqrt%283%2F4%29+
Hope this makes sense