The equation 2•x² + 4•y² + 8•x - 16•y + F = 0, at a value of F is the graph of an ellipse when we have;
a. A value of F that results in the equation of an ellipse is; F = 23
b. A point on the ellipse is ((-2), 2.5)
Which method can be used to find the value of F that gives the graph of an ellipse?
The equation of an ellipse can be presented as follows;
[tex] \frac{(x - h) ^{2} }{ {a}^{2} } + \frac{(y - k) ^{2} }{ {b}^{2} } = 1[/tex]
The given equation can be presented as follows;
2•x² + 4•y² + 8•x - 16•y + F = 0
Collecting like terms, gives;
2•(x² + 4•x) + 4•y² - 16•y + F = 0
Adding a constant value to both sides, we get;
2•(x² + 4•x + 4) + 4•y² - 16•y + 16 + F = 24
2•(x + 2)•(x + 2) + (2•y - 4)•(2•y - 4) + F = 24
2•(x + 2)² + 4•(y - 2)² + F = 24
2•(x + 2)² + 4•(y - 2)² + F - 23 = 24 - 23 = 1
When F = 23, we have;
2•(x + 2)² + 4•(y - 2)² + 23 - 23 = 1
- 2•(x + 2)² + 4•(y - 2)² = 1
Which gives;
[tex] \mathbf{\frac{(x + 2) ^{2} }{ \left( {\frac{1}{ \sqrt{2} }} \right)^{2}} + \frac{(y - 2) ^{2} }{ \left( {\frac{1}{ 2} } \right)^{2}} } = 1[/tex]
Where;
b) A point on the ellipse can be found as follows;
- 2•(x + 2)² + 4•(y - 2)² = 1
When x = -2
2•((-2) + 2)² + 4•(y - 2)² = 1
0 + 4•(y - 2)² = 1
y - 2 = √(1/4) = 1/2
A point on the ellipse is therefore;
When y = 2
2•(x + 2)² + 4•(2 - 2)² = 1
2•(x + 2)² + 0 = 1
2•(x + 2)² = 1
(x + 2)² = 1/2
x + 2 = √(1/2)
x = √(1/2) - 2
Another point on the ellipse is therefore;
((√(1/2) - 2), 0)
Learn more about the equation of an ellipse here:
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