Question

Find the equation of ellipse having eccentricity 3/8 and distance between its foci is 6.​

Answer 1

Answer:

The equation of the ellipse is: $\frac{x^{2}}{64}+\frac{y^{2}}{55}=1$

Step-by-step explanation:

It is provided that the eccentricity of the ellipse is, $e=\frac{3}{8}$

The distance between the foci is, $2ae=6$.

The equation of an ellipse is:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Compute the value of a as follows:

$2ae=6\\a=\frac{6}{2e}\\= \frac{6}{2}\times\frac{8}{3}\\=8\\a^{2}=64$

The value of b can be computed using the formula $b^{2}=a^{2}(1-e^{2})$.

The value of b is:

$b^{2}=a^{2}(1-e^{2})\\=8^{2}(1-(\frac{3}{8})^{2} )\\=55$

The equation of the ellipse is:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\\frac{x^{2}}{64}+\frac{y^{2}}{55}=1$

Answer 2

The correct answer is 8n the image provided by me.

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