Find the center, vertices, and foci of the ellipse 9x^2 + 4y^2 + 36x -8y + 4 =0  

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We start with writing the terms containing x and y together. We get 9x^2 + 36x + 4y^2 – 8y + 4 =0

=> 9(x^2 + 4x) + 4(y^2 – 2y) + 4 =0

Now complete the squares

=> 9(x^2 + 4x +4) +4(y^2 – 2y +1) = -4 + 36 + 4

divide both the sides by 36

=> (x^2 + 4x +4)/4 + (y^2 – 2y +1)/9 = 1

=> (x + 2) ^2 / 4 + (y – 1)/9 = 1

This is in the form (x – h) ^2/a^2 + (y – k) ^2/b^2 = 1

a = 3, b= 2, h = -2 and k=1.

The center of the ellipse is (h, k) or (-2, 1).

The vertices of the ellipse are (h, k-a) = (-2, -2) and (h, k + a) = (-2, 4)

The foci are the points (h, k + sqrt (a^2 + b^2) = (-2, 1 + sqrt 5) and (h, k- sqrt (a^2 + b^2) = (-2, 1- sqrt 5)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The equtaion of the ellipse is :

x^2/a^2 + y^2/b^2 = 1

We'll re-write the equation:

x^2*b^2 + y^2*a^2 - a^2*b^2 = 0

9x^2 + 4y^2 + 36x -8y + 4 =0

We'll complete the squares:

9x^2 + 36x + 36 - 36 + 4y^2 - 8y + 4 - 4 + 4 = 0

(x+6)^2 + (2y-2)^2 - 36 = 0

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

We'll divide by 36:

(x+6)^2/6^2 + (y-1)^2/3^2 = 1

semi-major axis is:a = 6

semi-minor axis is:b=3

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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The equation of the ellipse is :

9x^2 +4y^2+36x-8y+4 = 0

9x^2-36 +4y^2-8y = -4.

We add 6^2+2^2 to  both sides.

(3x)^2-36x + 6^2 +( 2y)^2-8y +2^2 =  -4+6^2+2^2.

(3x-6)^2 +(2y-2)^2 = 36 = 6^2.

(3x-6)^2/36 +(2y-2)^2/36 = 1.

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

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

Which could be compared with the standard ellipse x^2/b^2+y^2/a^2= 1

Clearly (1)  is an ellipse whose semi major axis is 3 on y axis and  with centre  (-2, 1) and  focus at (0,3sqrt(1-2^2/3^2) = (0,sqrt5) and (0 -sqrt5). The vertices at (0,3) and (0 -3).

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