How do I graph the hyperbola with the equation $4x^2−25y^2−50y−125=0$ ?

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Answer 1 · 2015-03-02T12:02:05

We have to write it in the standard form:

$4x^2-25y^2-50y-125=0rArr$

$4x^2-25(y^2+2y)=125rArr$

$4x^2-25(y^2+2y+1-1)=125rArr$

$4x^2-25(y+1)^2+25=125rArr$

$4x^2-25(y+1)^2=100rArr$

$4/100x^2-25/100(y+1)^2=1rArr$

$x^2/25-(y+1)^2/4=1$

This is an hyperbola centered in $C(0,-1)$ , with the branches left and right (there is an $1$ at the second member), with the asymptotes:

$m=+-4/25$ , so:

$y-(-1)=+-4/25(x-0)rArry=+-4/25x-1$ .

graph{4x^2-25y^2-50y-125=0 [-10, 10, -5, 5]}

How do I graph the hyperbola with the equation 4x^2−25y^2−50y−125=04x^2−25y^2−50y−125=0?