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Question 1169000: find the equation of the ellipse if the vertex is at (5,1) and foci at (-2,1) and (4,1). express your answer in general form.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
The given information about the ellipse is:
Vertex: $(5, 1)$
Foci: $(-2, 1)$ and $(4, 1)$
Since the y-coordinates of the vertex and the foci are the same, the major axis of the ellipse is horizontal.
The center of the ellipse is the midpoint of the foci:
Center $(h, k) = \left( \frac{-2 + 4}{2}, \frac{1 + 1}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)$
The distance between the center and a focus is $c$. Using the focus at $(4, 1)$:
$c = \sqrt{(4 - 1)^2 + (1 - 1)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3$
The distance between the center and a vertex is $a$. Using the vertex at $(5, 1)$:
$a = \sqrt{(5 - 1)^2 + (1 - 1)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$
We know the relationship between $a$, $b$, and $c$ for an ellipse: $c^2 = a^2 - b^2$.
Substituting the values of $a$ and $c$:
$3^2 = 4^2 - b^2$
$9 = 16 - b^2$
$b^2 = 16 - 9 = 7$
The standard form of the equation of a horizontal ellipse with center $(h, k)$ is:
$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
Substituting the values of $h=1$, $k=1$, $a^2 = 16$, and $b^2 = 7$:
$\frac{(x - 1)^2}{16} + \frac{(y - 1)^2}{7} = 1$
To express the equation in general form, we eliminate the denominators and expand:
$7(x - 1)^2 + 16(y - 1)^2 = 16 \times 7$
$7(x^2 - 2x + 1) + 16(y^2 - 2y + 1) = 112$
$7x^2 - 14x + 7 + 16y^2 - 32y + 16 = 112$
$7x^2 + 16y^2 - 14x - 32y + 23 = 112$
$7x^2 + 16y^2 - 14x - 32y + 23 - 112 = 0$
$7x^2 + 16y^2 - 14x - 32y - 89 = 0$
Final Answer: The final answer is $\boxed{7x^2 + 16y^2 - 14x - 32y - 89 = 0}$
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