Question
Find the equation of an ellipse whose vertices are $(0,\pm10)$ and eccentricity $\text{e}=\frac{4}{5}.$

Answer

Let the equation of the ellipse be

$\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ,\dots(\text{i})$

The coordinates of vertices are $(0,\pm\text{b})$ i.e., $(0,\pm10).$

$\therefore\ \text{b}=10$

$\Rightarrow\text{b}^2=100$

Now$\text{a}^2=\text{b}^2\big(1-\text{e}^2\big)$

$\Rightarrow\text{a}^2=100\Big[1-\Big(\frac{4}{5}\Big)^2\Big]$

$\Rightarrow\text{a}^2=100\Big[1-\frac{16}{25}\Big]$

$\Rightarrow\text{a}^2=100\Big[\frac{9}{25}\Big]$

$\Rightarrow\text{a}^2=4\times9=36$

Putting a2 = 36 and b2 = 100 in equation (i), we get

$\frac{\text{x}^2}{36}+\frac{\text{y}^2}{100}=1$

$\Rightarrow\frac{100\text{x}^2+36\text{y}^2}{3600}=1$

$\Rightarrow100\text{x}^2+36\text{y}^2=3600$

this is the equation of the required ellipse.

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