The eccentricity of an ellipse is $2/3$, latus rectum is $5$ and centre is $(0, 0)$. The equation of the ellipse is
The eccentricity of an ellipse is $2/3$, latus rectum is $5$ and centre is $(0, 0)$. The equation of the ellipse is
- A
$\frac{{{x^2}}}{{81}} + \frac{{{y^2}}}{{45}} = 1$
- B
$\frac{{4{x^2}}}{{81}} + \frac{{4{y^2}}}{{45}} = 1$
- C
$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1$
- D
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
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Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :
- [JEE MAIN 2024]