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The equation of an ellipse is given \(2x^2 - 8x + 3y^2 + 6y + 5 = 0.\) Find the maximum value of the \(x\)-coordinate of a point on this ellipse.

 Jul 29, 2019
 #1
avatar+6192 
+2

\(2x^2 - 8x + 3y^2+6y+5=0\\ 2(x^2-4x) + 3(y^2+2y)+5=0\\ 2(x^2-4x+4-4)+3(y^2+2y+1-1) + 5 = 0\\ 2(x-2)^2-8+3(y+1)^2-3+5=0\\ 2(x-2)^2+3(y+1)^2=3\\ \left(\dfrac{x-2}{\sqrt{\dfrac 3 2}}\right)^2+(y+1)^2 = 1\)

 

\(\text{There are no $xy$ terms so the ellipse isn't rotated at all, thus the maximum $x\\$ coordinate will occur when $y=0$}\\ x-2 = \sqrt{\dfrac 3 2 }\\ x = 2 + \sqrt{\dfrac 3 2}\)

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 Jul 29, 2019
 #2
avatar+111389 
+1

2x^2  - 8x + 3y^2  + 6y  + 5  = 0

 

2x^2 - 8x + 3y^2  + 6y  =  - 5       compete the square on x , y

 

2 ( x^2 - 4x + 4)  + 3(y^2 + 2y + 1)   =   -5  + 8 + 3

 

2 ( x - 2)^2  +  3 ( y + 1)^2   =  6           divide both sides by 6

 

(x - 2)^2          (y + 1)^2

_______  +     ________    =       1

      3                    2

 

We have the form

 

(x - h)^2          ( y - k)^2

________  +  _________  =  1

     a^2                  b^2

 

The center of this ellipse  is  ( 2, - 1)

 

And the major  axis lies along  x

 

And the max value of x  =   ( 2 + a)  =  (2 + √3)  ≈  3.732

 

Here is a graph that shows this :  https://www.desmos.com/calculator/qu9bzxtafp

 

 

cool cool cool

 Jul 29, 2019

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