Question 623675
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Hi, there--

The Problem:
What are the vertices of the ellipse given by the equation x2 + 4y2 + 10x – 56y + 205 = 0?

A Solution:
By using the complete the square method, we will put this equation into this form:
{{{((x-h)^2)/a^2+((y-k)^2)/b^2=1}}}

Then we can easily see find the coordinates of the focal points.

x^2 + 4y^2 + 10x – 56y + 205 = 0

Subtract 205 from both sides of the equation.
x^2 +4y^2 + 10x -56y = -205

Rearrange the terms on the left hand side to gather all x-terms, then all y-terms.
(x^2 + 10x) + (4y^2-56y) = -205

Factor a 4 out of the y terms so the the coefficient of the y-squared term will be 1.
(x^2 + 10x) + 4(y^2-14y) = -205

We need to add a constant term to (x^2 +10x) that will give us a perfect square polynomial. To 
find that constant, take half the coefficient of 10x and square it; half of 10 is 5, and 5-squared is 
25.

We will add 25 to both sides of the equation.
(x^2 + 10x + 25) + 4(y^2 - 14y) = -205 + 25

Now we need to add a value to (y^2-14x) that will give us a perfect square polynomial. To find 
the number, take half the coefficient of -14y and square it. So half of -14 is -7 and -7 squared 
is 49.

We will add 49 inside the parentheses on the left side of the equation. On the right side of the 
equation we need to add 4*49 since every term inside those parentheses is multiplied by 4.
(x^2 + 10x + 25) + 4(y^2 -14y + 49) = -205 + 25 + 4*49

Simplify the right hand side of the equation.
(x^2 + 10x + 25) + (4y^2 -56y + ) = -205 + 25 + 4*49

Since we now have two square polynomials, we can write them in factored form. 
(x+ 5)^2 + 4(y-7)^2 = 16
.
Finally divide both sides of the equation by 16.
{{{((x+5)^2)/16 +(4(x-7)^2)/16=16/16}}}

We can reduce 4/16 to 1/4 and 16/16 to 1.
{{{((x+5)^2)/16 +((x-7)^2)/4=1}}}

Represent the denominators as squares.
{{{((x-(-5))^2)/(4)^2 +((x-7)^2)/(2)^2=1}}}

After ALL this work, we have the equation in the form we want. 
The center of the ellipse is the point (h,k) = (-5,7)
The value of a is 4 and the value of b is 2.

The vertices are located at (h+a,k) and (h-a,k). In this equation, we have the two vertices at 
(-5+4,-7)=(-1,7) and (-5-4,7)=(-9,7)

The co-verticies are located at (h, k+b) and (h,k-b). In this equation, the co-verticies are located at 
(-5,7+2)=(-5,9) and (-5,7-2)=(-5,5)
There is a lot of algebra here. Feel free to email me if you have questions about any of the solution.

Ms.Figgy
math.in.the.vortex@gmail.com


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