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.
\(2x^2 - 8x + 3y^2+6y+5=0\\ 2(x^2-4x) + 3(y^2+2y)+5=0\\ 2(x^2-4x+4-4)+3(y^2+2y+1-1) + 5 = 0\\ 2(x-2)^2-8+3(y+1)^2-3+5=0\\ 2(x-2)^2+3(y+1)^2=3\\ \left(\dfrac{x-2}{\sqrt{\dfrac 3 2}}\right)^2+(y+1)^2 = 1\)
\(\text{There are no $xy$ terms so the ellipse isn't rotated at all, thus the maximum $x\\$ coordinate will occur when $y=0$}\\ x-2 = \sqrt{\dfrac 3 2 }\\ x = 2 + \sqrt{\dfrac 3 2}\)
.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