9x^2+4y^2+36x-24y+36=0

Question

Find the center, foci, and vertices of the ellipse

Answer 1

9x^2+4y^2+36x-24y+36 = 0

add 36  to each side.

9x^2+4y^2+36x-24y+36+36 = 36

9x^2+36x+36+4y^2-24y+36 = 36

9(x^2+4x+4)+4(y^2-6y+9) = 36

Divide to 36 each side.

9(x^2+4x+4)/36+4(y^2-6y+9)/36 = 36/36

(x^2+4x+4)/4+(y^2-6y+9)/9 = 1

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

If the large denominator is under the y term , then the ellipse is vertical.

Compare the standard form of ellipse is (x-h)^2/b^2+(y-k)^2/a^2 = 1

Here a = 3,b = 2, h = -2,k = 3

Center is (h,k) = (-2,3)

Vertices are (h,k+a)(h,k-a)

(h,k+a) = (-2,3+3) = (-2,6)

(h,k-a) = (-2,3-3) = (-2,0)

c = √(a^2-b^2) = √9-4 =√5

Now fnd foci is (h,k+c) ,(h,k-c)

(h,k+c) = (-2,3+√5)

(h,k-c) = (-2,3-√5)