How do you find the center, vertices, and foci of an ellipse #(x-3)^2 / 16 + (y+2)^2 / 9=1#?

3 Answers
Nov 22, 2016

Answer:

Please see the explanation.

Explanation:

The standard form for the equation of an ellipse is:

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

The center is: #(h,k)#
The vertices on the major axis are: #(h - a, k) and (h + a, k)#
The vertices on the minor axis are: #(h, k -b) and (h, k + b)#
The foci are: #(h - sqrt(a^2 - b^2), k) and (h + sqrt(a^2 - b^2), k)#

To put the given equation in standard form, change the + 2 to - -2 and write the denominators as squares:

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

The center is: #(3,-2)#
The vertices on the major axis are: #(-1, -2) and (7, -2)#
The vertices on the minor axis are: #(3, -5) and (3, 1)#

Evaluate: #sqrt(a^2 - b^2) = sqrt(4^2 - 3^2) = sqrt(16 - 9) = sqrt(5)#

The foci are: #(3 - sqrt(5), -2) and (3 + sqrt(5), -2)#

Nov 22, 2016

Answer:

The center is #=(3,-2)#
The vertices are #(7,-2)#, #(-1,-2)#, #(3,1)#, #(3,-5)#
The foci are F#(3-sqrt7,-2)# and F'#(3+sqrt7,-2)#

Explanation:

We compare this equation to

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

The center is #(h,k)=(3,-2)#

The vertices are A #(h+a,k)=(7,-2)# and A'#(h-a,k)=(-1,-2)#
and

B #(h,k+b)=(3,1)# and B' #(h,,k-b)=(3,-5)#

To calculate the foci, we need c

#c^2=a^2-b^2=16-9=7#

The foci are F #(h-sqrtc,k)=(3-sqrt7,-2)#
and F' #(h+sqrtc,k)=(3+sqrt7,-2)#

graph{(x-3)^2/16+(y+2)^2/9=1 [-8.56, 11.44, -8.316, 1.685]}

Nov 22, 2016

Answer:

Centre is (3, -2), focii are #(-sqrt7 +3, -2) and (sqrt7 +3, -2)#. vertices (on horizontal axis) would be at (-4+3,-2) and (4+3,-2) Or (-1,-2) and (7,-2)

Explanation:

Standard equation of an ellipse centered at (h,k) is #(x-h)^2 / a^2 + (y-k)^2 /b^2 =1# with major axis 2a and minor axis 2b.

The foci of this ellipse are at (c+h, k) and (-c+h, k). The vertices on horizontal axis would be at (-a+h,k) and (a+h,k), where #c^2= a^2 -b^2#

Comparing the given equation with the standard one, it is seen that a=4, b=3, c=#sqrt(4^2-3^2)= sqrt 7#. Also h= 3, k=-2

Hence Centre is (3, -2), focii are #(-sqrt7 +3, -2) and (sqrt7 +3, -2)#. vertices (on horizontal axis) would be at (-4+3,-2) and (4+3,-2) Or (-1,-2) and (7,-2).