SOLUTION: (x^2/25) - (y^2/b^2) =1 Find the foci, directrices, and eccentricity of hyperbola. I am struggling with the method for this question, I understand that the value a is 5, but

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: (x^2/25) - (y^2/b^2) =1 Find the foci, directrices, and eccentricity of hyperbola. I am struggling with the method for this question, I understand that the value a is 5, but       Log On


   

Question 608068: (x^2/25) - (y^2/b^2) =1
Find the foci, directrices, and eccentricity of hyperbola.
I am struggling with the method for this question, I understand that the value a is 5, but I am unsure on how to use this to get an answer???
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
(x^2/25) - (y^2/b^2) =1
Find the foci, directrices, and eccentricity of hyperbola.
**
Given equation is that of a hyperbola with horizontal transverse axis. (x-term to the left of y-term)
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
..
For given equation:
a^2=25
a=5
But there is not sufficient information to solve:
Foci (c)=√(a^2+b^2)
Eccentricity (e)=c/a
or
Directrices(ħa/e)
The value of b is required to get numerical answers to these problems