How do you find the center, vertices, foci and asymptotes of #-4x^2 +25y^2 -8x+ 150y+ 121= 0#?

1 Answer
Jan 19, 2017

See explanation. See the hyperbola, center, asymptotes, axes and vertices, in the Socratic graph..


The form #25y^2-4x^2 for the second degree terms suggest that I

can first find asymptotes #A1 xx A2=0# and center C, their point of


Let the equation to this hyperbola be

#A1 xx A2 = (5y-2x+ c_1)(5y+2x+c_2)=c#

By comparison with constant and first degree terms,

#c_1=13, c_2=17 and c = 100#

The asymptotes are given by

#A1xxA2=(5y-2x+13)(5y+2y+17)=0#m meeting athe center of the

hyperbola C(1, -3).

Note that the bisectors through the center parallel to the axes,

x =- 1 and y = -3 are the axes of the hyperbola.

At #x =-1, y = -5 and -1. So, major axis 2a = (-1)-(-5)=4. And so, a = 2.

The vertices are #A (-1, -1) and A' (-1, -5)#.

The Socratic asymptotes-center-hyperbola graph is for this form of

the equation.

Only foci wait for entry, in my next edition,

. graph{((5y-2x+13)(5y+2x+17)-100)(5y-2x+13)(5y+2x+17)(y+3)(x+1)=0 [-20, 20, -10, 10]}