How do you find the center, vertices, foci and eccentricity of 9x^2+4y^2-36x+8y+31=0?

1 Answer
Aug 13, 2016

Centre: (2,-1)

Vertices: (2, 1/2) and (2,-5/2)

Co-Vertices: (1,-1) and (3,-1)

Foci: (2, (-2+sqrt(5))/2) and (2,(-2-sqrt(5))/2)

Eccentricity: sqrt(5)/3

Explanation:

The technique we want to use is called completing the square. We shall use it on the x terms first and then the y.

Rearrange to

9x^2 + 4y^2 - 36x + 8y = -31

Focussing on x, divide through by the x^2 coefficient and add the square of half the coefficient of the x^1 term to both sides:

x^2 + 4/9y^2 - 4x + 8/9y +(-2)^2 = -31/9 + (-2)^2

(x-2)^2 + 4/9y^2 + 8/9y = 5/9

Divide through by y^2 coefficient and add square of half the coefficient of the y^1 term to both sides:

9/4(x-2)^2 + y^2 + 2y + (1)^2= 5/4+ (1)^2

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

Divide by 9/4 to simplify:

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

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

General equation is

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

where (a,b) is the centre and h, k are the semi-minor/major axis.

Reading off the centre gives (2, -1).

In this case, the y direction has a bigger value than the x, so the ellipse will be stretched in the y direction. k^2 > h^2

The vertices are obtained by moving up the major axis from the centre. Ie +-sqrt(k) added to the y coordinate of the centre.

This gives (2, 1/2) and (2, -5/2).

The co-vertices lie on the minor axis. We add +-sqrt(h) to the centre's x coordinate to find these.

(1,-1) and (3,-1)

Now, to find the foci:

c^2 = k^2 - h^2

c^2 = 9/4 - 1

c^2 = 5/4 implies c = +-sqrt(5)/2

Foci will be situated along the line x = 2 at +-sqrt(5)/2 from y = -1.

therefore foci at (2, (-2+sqrt(5))/2) and (2,(-2-sqrt(5))/2)

Finally the eccentricity is found using

e=sqrt(1-h^2/k^2)

e=sqrt(1-1/(9/4)) = sqrt(1-4/9) = sqrt(5)/3