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?

1 Answer
Mar 15, 2017

Answer:

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#