SOLUTION: Hello, could you please help me figure out where I am going wrong with the foci and vertices of {{{ 25x^2-4y^2+50x+16y= 91 }}}. I keep getting an "a" of 2 and a "b" of 5 (with "c"

Question 545573: Hello, could you please help me figure out where I am going wrong with the foci and vertices of . I keep getting an "a" of 2 and a "b" of 5 (with "c" being sqrt( 29 ) ) but my calculator's graph says it should be different (an "a" of 1/2). Thank you.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find Foci and Vertices
25x^2-4y^2+50x+16y= 91
**
25x^2+50x+-4y^2+16y= 91
complete the square
25(x^2+2x+1)-4(y^2-4y+4)= 91+25-16=100
25(x+1)^2-4(y-2)^2=100
divide by 100
(x+1)^2/4-(y-2)^2/25=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1
For given equation:
center:(-1,2)
..
a^2=4
a=2
vertices: (-1�a, 2)=(-1�2,2)=(-3,2) and (1,2)
..
b^2=25
b=5
..
c^2=a^2+b^2=4+25=29
c=√29≈5.4
Foci: (-1�√29,2), (-1�6.4,2)=(-7.4,2) and (5.4,2)
Seems like you got a, b, and c correct. Maybe you did not punched the right data into your graphing calculator.
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