Question

How do you put `12x^2+3y^2-30y+39=9` in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Answer 1

`(4x^2)/15 + (y-5)^2/15 = 1; " center": (0, 5)`
vertices: `(0, 5+sqrt(15)), (0, 5- sqrt(15))`
endpoints (co-vertices): `(-sqrt(15)/2, 5), (sqrt(15)/2, 5)`
foci: `(0, 5+(3sqrt(5))/2), (0, 5-(3sqrt(5))/2)`; `e = sqrt(3)/2`

Explanation:

Given: `12x^2 + 3y^2-30y+39 = 9`

To Find the standard form of the equation , use the completing of the square technique :

Group both the `x`-terms and the `y`-terms together on the left-side of the equation and all the other constants on the right side:
`12x^2 + (3y^2-30y) = -30`

Factor if needed: `" "12x^2 + 3(y^2-10y) = -30`

We only need to use completing of the square with the `y`-terms. Half the `-10y " term "-> -5 " and add the " 3(-5)^2` to the right side that was added to the left side when we completed the square:

`" "12x^2 + 3(y-5)^2 = -30 + 3(5)^2`

`" "12x^2 + 3(y-5)^2 = 45`

Divide everything by `45` to make the equation `= 1`:

Standard form of the equation:

`" "(4x^2)/15 + ((y-5)^2)/15 = 1 " or " x^2/(sqrt(15)/2)^2 + (y-5)^2/(sqrt(15)^2) = 1`

Because we have a "sum" in the equation, we have an ellipse . In an ellipse `a` is always the major axis (longest). This means we have a Vertical Major Axis ellipse: `(x-h)^2/b^2 + (y-k)^2/a^2 = 1`

Where `" **center** "= (h, k) = (0, 5)`

`a = sqrt(15), " "b = sqrt(15)/2`
`c = sqrt(a^2 - b^2) = sqrt(15-15/4) = sqrt(45/4) = (3sqrt(5))/2`

Vertices: `(h, k+a), (h, k-a): (0, 5 +sqrt(15)), (0, 5-sqrt(15))`

Endpoints (co-vertices): `(h-b, k), (h+b, k): (-sqrt(15)/2, 5), (sqrt(15)/2, 5)`

Foci: `(h, k+c), (h, k-c): (0, 5+(3sqrt(5))/2), (0, 5-(3sqrt(5))/2)`

Eccentricity:
`varepsilon = c/a = ((3sqrt(5))/2)/sqrt(15)=(3sqrt(5))/2 * 1/sqrt(15) *sqrt(15)/sqrt(15) = (3sqrt(75))/30 = sqrt(75)/10`

`varepsilon =sqrt(3)/2`