How do you find the vertices and foci of #\frac { ( x + 2) ^ { 2} } { 16} - \frac { ( y - 2) ^ { 2} } { 9} = 1#?

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Answer 1 · 2018-03-12T12:30:06

Vertices are at # (-6,2) and (2,2)#
Focii are at # (-7,2) and (3,2)#

The standard form of equation of hyperbola with centre #(h,k)# is

#(x-h)^2/a^2-(y-k)^2/b^2=1 #.Transverse axis is horizontal. In

#(x+2)^2/4^2-(y-2)^2/3^2=1; h=-2 , k=2 , a=4 , b=3# ;

Centre is #(h,k) or (-2,2) # .The vertices are #a# units from the

center. Therefore vertices are at #(-2-4),2) and (-2+4),2)#

or at # (-6,2) and(2,2)# .The foci are #c# units from the

center. Moreover, #c^ 2=a^2+b^2 : c^2 = 16+9 or c = +- 5#

Therefore focii are at #(-2-5),2) and (-2+5),2)# or at

# (-7,2) and(3,2)#

graph{(x+2)^2/16-(y-2)^2/9=1 [-10, 10, -5, 5]} [Ans]