The equation of an ellipse with center (-9, 3), focus (-6, 3) and major axis length of 14 units is [tex]\frac{(x+9)^2}{49} + \frac{(y-3)^2}{40}= 1[/tex]
The standard form of the equation of an ellipse is:
[tex]\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}= 1[/tex]
The center of the ellipse, (h, h) = (-9, 3)
That is, h = -9, k = 3
The length of the major axis = 14 units
The length of the major axis = 2a
2a = 14
a = 14/2
a = 7
a² = 7²
a² = 49
The coordinates of the foci are [tex](h \pm c, k)[/tex]
One of the foci = (-6, 3)
h - c = -6
-9 - c = -6
c = -9 + 6
c = -3
[tex]c^2 = a^2 - b^2\\(-3)^2 = 7^2 - b^2\\9 = 49 - b^2\\b^2 = 49 - 9\\b^2 = 40[/tex]
Substitute a² = 49, b² = 40, h = -9, and k = 3 into the equation [tex]\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2}= 1[/tex]
[tex]\frac{(x+9)^2}{49} + \frac{(y-3)^2}{40}= 1[/tex]
The equation of an ellipse with center (-9, 3), focus (-6, 3) and major axis length of 14 units is [tex]\frac{(x+9)^2}{49} + \frac{(y-3)^2}{40}= 1[/tex]
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