How do you find the center, foci and vertices of #(x-6)^2/4+(y+7)^2/16=1#?

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Answer 1 · 2016-12-20T16:17:46

The center is #=(6,-7)#
The foci are #=(6,-7+2sqrt3)# and #=(6,-7-2sqrt3)#
The vertices are #(6,-3)#, #(6,-11)#, #(8,-7)# and #(4,-7)#

The equation represents an ellipse.

#(x-6)^2/4+(y+7)^2/16=1#

with the major axis vertical

The general equation is

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

The center of the ellipse is #(h,k)=(6,-7)#

Let's calculate #c=sqrt(b^2-a^2)=sqrt(16-4)=sqrt12=2sqrt3#

The foci are F#=(h,k+c)=(6,-7+2sqrt3)#

and F'#=(h,k-c)=(6,-7-2sqrt3)#

The vertices are

A#=(h,k+a)=(6,-7+4)=(6,-3)#

A'#=(h,k-a)=(6,-7-4)=(6,-11)#

B#=(h+b,k)=(6+2, -7)=(8,-7)#

and B'#=(h-b,k,)=(6-2,-7)=(4,-7)#

graph{(x-6)^2/4+(y+7)^2/16=1 [-10.13, 15.18, -11.41, 1.25]}