Given the equation of a hyperpola #(x+2)^2/a-(y-1)^2/16=1# find the coordinates of c,the foci and the vertices?

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Answer 1 · 2017-11-12T13:38:09

The center is: #(-2,1)#

The vertices are: #(-2-sqrta,1)# and #(-2+sqrta,1)#

The foci are: #(-2-sqrt(a-16),1)# and #(-2+sqrt(a-16),1)#

Given: #(x+2)^2/a-(y-1)^2/16=1" [1]"#

We need to make a few minor changes, to equation [1], so that it is in the standard form of equation [2]:

#(x - h)^2/a^2-(y-k)^2/b^2 = 1" [2]"#

Please observe that we wrote #+2# as #-(-2)# and wrote the denominators as squares.

#(x-(-2))^2/(sqrta)^2-(y-1)^2/4^2=1" [1.1]"#

We know that the center of equation [2] is #(h,k)#.

Matching this with equation [1.1], we observe that the center is: #(-2,1)#

We know that the vertices of equation [2] are #(h-a,k)# and #(h+a,k)#.

Matching this with equation [1.1] we observe that the vertices are: #(-2-sqrta,1)# and #(-2+sqrta,1)#

We know that the foci of equation [2] are #(h-sqrt(a^2-b^2),k)# and #(h+sqrt(a^2-b^2),k)#.

Matching this with equation [1.1] we observe that the foci are: #(-2-sqrt(a-16),1)# and #(-2+sqrt(a-16),1)#