How do you find the vertex, focus, and directrix of a hyperbola #((x+3)^2/4) - ((y-4)^2/25)=1#?

Question

2759 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T05:45:12

Vertices are #(-1,4)# and #(-5,4)#; focii are #(-3+sqrt29,4)# and #(-3-sqrt29,4)# and directrix are #x=-4/sqrt29# and #x=4/sqrt29#

Here the equation of hyperbola is

#(x+3)^2/4-(y-4)^2/25=1# or #(x+3)^2/2^2-(y-4)^2/5^2=1#

If the general equation of hyperbola is

#(x-h)^2/a^2-(y-k)^2/b^2=1#, then

vertices are #(h+-a,k)#, focii are #(h+-c,k)#, where #c^2=a^2+b^2#

and directrix are #x=+-a^2/c#

Here #a=2#, #b=5#, #c=sqrt(2^2+5^2)=sqrt29# and #h=-3#, #k=4#

Hence for hyperbola #(x+3)^2/2^2-(y-4)^2/5^2=1#

Vertices are #(-3+-2,4)# i.e. #(-1,4)# and #(-5,4)#

Focii are #(-3+sqrt29,4)# and #(-3-sqrt29,4)#

and directrix are #x=-4/sqrt29# and #x=4/sqrt29#
graph{(x+3)^2/2^2-(y-4)^2/5^2=1 [-24.42, 15.58, -5.68, 14.32]}