Determine the coordinates of the end points of latus rectum of a standard vertical ellipse.

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Solution

Given: A standard vertical ellipse.

The standard equation is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ where $b > a$ .

Now, we know that, the coordinates of the foci of a standard vertical ellipse are $(0, \pm b e)$ .

Hence, we have the y - coordinate of the end points of latus rectum as they lie on the same line as the foci.

Substituting the y value in the standard equation.

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

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

$\Rightarrow \frac{x^{2}}{a^{2}} = 1 - e^{2}$

We know that, the eccentricity of an ellipse with $b > a$ is given by $e^{2} = 1 - \frac{a^{2}}{b^{2}}$

$\Rightarrow \frac{x^{2}}{a^{2}} = \frac{a^{2}}{b^{2}}$

$\Rightarrow x^{2} = \frac{a^{4}}{b^{2}}$

$\Rightarrow x = \pm \frac{a^{2}}{b}$

Hence, the coordinates of the end points of latus rectum will be $(\frac{a^{2}}{b}, b e), (- \frac{a^{2}}{b}, b e), (- \frac{a^{2}}{b}, - b e)$ and $(\frac{a^{2}}{b}, - b e)$ .

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