Question

The extremities of the latus rectum of an ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

• A. $(\pm \sqrt{7},\pm \frac{9}{4})$
• B. $(\pm \sqrt{9},\pm \frac{16}{5})$
• C. $(\pm 4,\pm \frac{25}{4})$
• D. $(\pm 6,\pm \frac{4}{5})$

Answer 1

The correct option is A $(\pm \sqrt{7},\pm \frac{9}{4})$

Given ellipse is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Where $a=4,b=3$
Clearly $a>b$
$\Rightarrow e^2=1-\frac{b^2}{a^2}=\frac{7}{16}$
$\therefore e=\frac{\sqrt{7}}{4}$
Coordinates of extremities of latus rectum are given by
$(\pm ae,\pm \frac{b^2}{a})\equiv (\pm \sqrt{7},\pm \frac{9}{4})$