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

Jan 19, 2017

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

#### Explanation:

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

can first find asymptotes $A 1 \times A 2 = 0$ and center C, their point of

intersection.

Let the equation to this hyperbola be

$A 1 \times A 2 = \left(5 y - 2 x + {c}_{1}\right) \left(5 y + 2 x + {c}_{2}\right) = c$

By comparison with constant and first degree terms,

${c}_{1} = 13 , {c}_{2} = 17 \mathmr{and} c = 100$

The asymptotes are given by

$A 1 \times A 2 = \left(5 y - 2 x + 13\right) \left(5 y + 2 y + 17\right) = 0$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 $A \left(- 1 , - 1\right) \mathmr{and} A ' \left(- 1 , - 5\right)$.

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]}