Question

The eccentricity of the ellipse $a{x}^2+b{y}^2+2fx+2gy+c=0$ if axis of ellipse parallel to x-axis is

• A. $\left(\sqrt{\frac{b-a}{b}}\right)$
• B. $\left(\sqrt{\frac{a+b}{b}}\right)$
• C. $\left(\sqrt{\frac{a+b}{a}}\right)$
• D. none of these

Answer 1

The correct option is A

$\left(\sqrt{\frac{b-a}{b}}\right)$

$a{x}^2+b{y}^2+2fx+2gy+c=0a\{x^2+\frac{2fx}{a}\}+b\{y^2+\frac{2gy}{b}\}+c=0\Rightarrow a{(x+\frac{f}{a})}^2+b{(y+\frac{g}{b})}^2=(\frac{f^2}{a}+\frac{g^2}{b}-c)\Rightarrow \frac{{(x+\frac{f}{a})}^2}{\frac{(\frac{f^2}{a}+\frac{g^2}{b}-c)}{a}}+\frac{{(y+\frac{g}{b})}^2}{\frac{(\frac{f^2}{a}+\frac{g^2}{b}-c)}{b}}=1$
if e eccentricity, then
$\frac{(\frac{f^2}{a}+\frac{g^2}{b}-c)}{b}=\frac{(\frac{f^2}{a}+\frac{g^2}{b}-c)}{a}(1-e^2)\Rightarrow 1-e^2=\frac{a}{b}\Rightarrow e^2=\frac{b-a}{b}\therefore e=\sqrt{\left(\frac{b-a}{b}\right)}$