Question
Find the equation of an ellipse whose vertices are $(0,\pm10)$ and eccentricity $\text{e}=\frac{4}{5}.$
$\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ,\dots(\text{i})$
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.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(-\text{x})^{-1}$
$\sqrt{\tan\text{x}}$