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Determine the equation of the ellipse with the following properties: center at point (1,4), a focus at point (5,4) and eccentricity e=2/3

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Answer to a math question Determine the equation of the ellipse with the following properties: center at point (1,4), a focus at point (5,4) and eccentricity e=2/3

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Corbin
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106 Answers
Para resolver este problema, siga os passos abaixo:

1. Identifique as coordenadas do centro e do foco da elipse. O centro é \((1, 4)\) e o foco é \((5, 4)\).

2. Determine a distância do centro ao foco, que é \(c\). A distância entre \((1, 4)\) e \((5, 4)\) é:

c = \sqrt{(5 - 1)^2 + (4 - 4)^2} = \sqrt{4^2} = 4

3. Use a excentricidade para encontrar \(a\), que é o semi-eixo maior. A excentricidade é dada por:

e = \frac{c}{a} = \frac{2}{3}

Logo, temos:

a = \frac{c}{e} = \frac{4}{\frac{2}{3}} = 4 \cdot \frac{3}{2} = 6

4. Encontre \(b\), o semi-eixo menor, usando a relação \(c^2 = a^2 - b^2\). Substituindo os valores conhecidos:

4^2 = 6^2 - b^2

16 = 36 - b^2

b^2 = 36 - 16

b^2 = 20

b = \sqrt{20} = 2\sqrt{5}

5. A equação padrão de uma elipse centrada em \( (h, k) \) com semi-eixo maior \(a\) e semi-eixo menor \(b\) é:

\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1

Substituindo os valores encontrados:

\frac{(x-1)^2}{6^2} + \frac{(y-4)^2}{(\sqrt{20})^2} = 1

\frac{(x-1)^2}{36} + \frac{(y-4)^2}{20} = 1

Assim, a equação é:

\frac{(x-1)^2}{36}+\frac{(y-4)^2}{20}=1

Portanto, a resposta é:

\frac{(x-1)^2}{36}+\frac{(y-4)^2}{20}=1

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