How do you find all the critical points to graph #x^2/25 - y^2/49=1# including vertices?

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Answer 1 · 2018-04-15T01:03:36

Below

general form of a hyperbola: #x^2/a^2-y^2/b^2=1#

#a^2=25# so #a=5#

#b^2=49# so #b=7#

the eccentricity is equal to #b^2=a^2(e^2-1)#

#49=25(e^2-1)#

#49/25+1=e^2#

#74/25=e^2#

#e=sqrt74/5#

Focus #(+-ae,0)#

#(+-5timessqrt74/5,0)#

#(+-sqrt74,0)#

Directrix: #x=a/e# SO #x=5/(sqrt74/5)# SO #x=25/sqrt74#

Asymptote: #y=b/ax# SO #y=7/5x#

Vertices: #(+-5,0)#