Question

The center of an ellipse is located at (0, 0). One focus is located at (12, 0), and one directrix is at x = 14 1/12

Answer 1

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Answer 2

The equation of the ellipse is $\frac{x^2}{169}+\frac{y^2}{25}=1$.

Correct Question:

The center of an ellipse is located at (0, 0). One focus is located at (12, 0), and one directrix is at x = 14 1/12. Find the equation of the ellipse.

Given:

The center of an ellipse is located at (0, 0). One focus is located at (12, 0), and one directrix is at $x=14\frac{1}{12}$.

To Find:

The standard equation of an ellipse is given by $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$. The center of the ellipse is at (0,0) so, h=0 and k=0.

The focus of an ellipse is given by $(c,0)$. So $c=12$.

The equation of directrix is given by $x=\frac{a^2}{c}$.

On comparing the given equation of directrix with the above-mentioned equation,

$\frac{a^2}{c}=14\frac{1}{12}\\ \frac{a^2}{c}=\frac{169}{12}$

Substitute the value c=12 and find a.

$\frac{a^2}{12}=\frac{169}{12}\\a^2=169\\a=13$

Use the equation $c^2=a^2-b^2$ and find b.

$(12)^2=(\frac{13}{12}) ^2-b^2\\b^2=169-144\\ =25\\b=5$

Substitute the value of a and b in the standard equation of the ellipse.

$\frac{(x-0)^2}{13^2}+\frac{(y-0)^2}{5^2}=1\\\frac{x^2}{169}+\frac{y^2}{25}=1$

Thus, the equation of the ellipse is $\frac{x^2}{169}+\frac{y^2}{25}=1$.