Question

How do you put `6x^2+5y^2-24x+20y+14=0` in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Answer 1

Standard forms for the equation of an ellipse are `(x-h)^2/a^2+(y-k)^2/b^2=1` or `(y-k)^2/a^2+(x-h)^2/b^2=1; a > b`

Explanation:

Reference for an ellipse
Given: `6x^2+5y^2-24x+20y+14=0`

Add `6h^2 + 5k^2 - 14` to both sides of the equation:

`6x^2-24x+6h^2+5y^2+20y+5k^2=6h^2+5k^2-14`

Remove a common factor of 6 from the first 3 terms and a common factor of 5 from the next 3 terms:

`6(x^2-4x+h^2)+5(y^2+4y+k^2)=6h^2+5k^2-14" [1]"`

To find the value of h, set the middle term of the right side of the pattern `(x - h)^2 = x^2 - 2hx+h^2` equal to the middle term in the first parenthesis in equation [1]:

`-2hx = -4x`

`h = 2`

Substitute the left side of the pattern into equation [1]:

`6(x - h)^2+5(y^2+4y+k^2)=6h^2+5k^2-14" [2]"`

Substitute 2 for h everywhere in equation [2]:

`6(x - 2)^2+5(y^2+4y+k^2)=6(2)^2+5k^2-14" [3]"`

To find the value of k, set the middle term of the right side of the pattern `(y - k)^2 = y^2 - 2ky+k^2` equal to the middle term in the second parenthesis in equation [3]:

`-2ky = 4y`

`k = -2`

Substitute the left side of the pattern into equation [3]:

`6(x - 2)^2+5(y - k)^2=6(2)^2+5k^2-14" [4]"`

Substitute -2 for k everywhere in equation [4]:

`6(x - 2)^2+5(y - -2)^2=6(2)^2+5(-2)^2-14" [5]"`

Simplify the right side of equation [5]:

`6(x - 2)^2+5(y - -2)^2=30" [6]"`

Divide both sides of the equation by 30:

`(x - 2)^2/5+(y - -2)^2/6=1" [7]"`

Make the denominators squares and swap terms:

`(y - -2)^2/(sqrt(6))^2+(x - 2)^2/(sqrt5)^2=1" [8]"`

Equation [8] is the standard form where, `h = 2, k = -2, a = sqrt(6), and b = sqrt(5)`

From the reference:

`c = sqrt(a^2-b^2)`

`c = sqrt(6 - 5)`

`c = 1`

The center is at `(h,k)`:

`(2,-2)`

The vertices are at #(h, k-a) and (h,k+a):

`(2,-2-sqrt(6)) and (2,-2+sqrt(6))`

The foci are at: #(h, k-c) and (h,k+c):

`(2,-2-1) and (2, -2+1)`

`(2,-3) and (2,-1)`

The co-vertices are at `(h-b, k) and (h+b,k)`:

`(2-sqrt(5),-2) and (2+sqrt(5),-2)`

The eccentricity is `c/a = 1/sqrt(6)= sqrt(6)/6`