How do you find the center, vertices, and foci of an ellipse (x-3)^2 / 16 + (y+2)^2 / 9=1(x3)216+(y+2)29=1?

3 Answers
Nov 22, 2016

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(xh)2a2+(yk)2b2=1

The center is: (h,k)(h,k)
The vertices on the major axis are: (h - a, k) and (h + a, k)(ha,k)and(h+a,k)
The vertices on the minor axis are: (h, k -b) and (h, k + b)(h,kb)and(h,k+b)
The foci are: (h - sqrt(a^2 - b^2), k) and (h + sqrt(a^2 - b^2), k)(ha2b2,k)and(h+a2b2,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(x3)242+(y2)232=1

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

Evaluate: sqrt(a^2 - b^2) = sqrt(4^2 - 3^2) = sqrt(16 - 9) = sqrt(5)a2b2=4232=169=5

The foci are: (3 - sqrt(5), -2) and (3 + sqrt(5), -2)(35,2)and(3+5,2)

Nov 22, 2016

The center is =(3,-2)=(3,2)
The vertices are (7,-2)(7,2), (-1,-2)(1,2), (3,1)(3,1), (3,-5)(3,5)
The foci are F(3-sqrt7,-2)(37,2) and F'(3+sqrt7,-2)(3+7,2)

Explanation:

We compare this equation to

(x-h)^2/a^2+(y-k)^2/b^2=1(xh)2a2+(yk)2b2=1

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

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

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

To calculate the foci, we need c

c^2=a^2-b^2=16-9=7c2=a2b2=169=7

The foci are F (h-sqrtc,k)=(3-sqrt7,-2)(hc,k)=(37,2)
and F' (h+sqrtc,k)=(3+sqrt7,-2)(h+c,k)=(3+7,2)

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

Nov 22, 2016

Centre is (3, -2), focii are (-sqrt7 +3, -2) and (sqrt7 +3, -2)(7+3,2)and(7+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(xh)2a2+(yk)2b2=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^2c2=a2b2

Comparing the given equation with the standard one, it is seen that a=4, b=3, c=sqrt(4^2-3^2)= sqrt 74232=7. Also h= 3, k=-2

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