Question

If the equation $\frac{x^2}{9-c}+\frac{y^2}{5-c}=1$ represents an ellipse, then the foci are :

• A. $(\pm 3,0)$
• B. $(\pm 2,3)$
• C. $(\pm 4,0)$
• D. $(\pm 2,1)$
• E. $(\pm 2,0)$

Answer 1

The correct option is E $(\pm 2,0)$

Given that the equation $\frac{x^2}{9-c}+\frac{y^2}{5-c}=1$ represents an ellipse.
Here, $a^2=9-c,b^2=5-c,a>b$
Therefore, eccenticity, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{5-c}{9-c}}$
$=\sqrt{\frac{4}{9-c}}=\frac{2}{\sqrt{9-c}}$
Foci $=(\pm ae,0)$

$=(\pm \sqrt{a-c}\times \frac{2}{\sqrt{a-c}},0)$
$=(\pm 2,0)$