How do you write the standard form of the hyperbola #-16x^2+9y^2+32x+144y-16=0#?

Question

3621 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-17T11:47:18

We can rewrite this as follows

#16x^2-9y^2-32x-144y+16=0# (Multiply with #-1# both sides)

#16*(x^2-2x)-9(y^2+16y)+16=0#

#16*(x^2-2x+1)-9(y^2+16+64)+(16-16+576)=0#

#16*(x-1)^2-9*(y+8)^2=-576#

#(y+8)^2/64-(x-1)^2/36=1#

#((y+8)/8)^2-((x-1)/6)^2=1#

Finally the standard form for the hyperbola is

#(y-(-8))^2/8^2-(x-1)^2/6^2=1#

If we plot it on the cartesian plane it is

enter image source here