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

Question

3990 views around the world

You can reuse this answer
Creative Commons License

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]}