Question

can someone help with this question: find the vertices of the hyperbola given by the equation:
16x^2-y^2+96x+10y+103

Answer 1

=0 at the end

Answer 2

Factor 16 from the x variables; subtract 103 to the RHS; complete the square separately on each variable; factor each variable expression as a perfect square trinomial to get an equation that looks like: 16(x+3)^2-(y-5)^2=16. The vertex is (-3,5).

Answer 3

so, would the verticies be at (-2,5) and (-4,5)?

Answer 4

Translate to standard position at (0,0) with (-3,5) , set x= 0 and then translate back again..

Answer 5

can you check this one for me: http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e4fec330b8b958804a8d59d

Answer 6

The vertices are on the transverse axis y=5, thus you would +/- 1 to the x-coordinate of the center to the vertices, which are as you stated. FYI, the foci are also on the TA, but they are sqrt(1+16)=sqrt(17) units from the center.

Answer 7

so i am correct.

Answer 8

on the vertices

Answer 9

can you just confirm the one about the foci?

Answer 10

Why don't u just use Wolfram if u just wan't to check yourself

http://www.wolframalpha.com/input/?i=16%28x%2B3%29^2-%28y-5%29^2%3D16

Click on properties.