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Question: The eccentricity of the ellipse\(ax^{2} + by^{2} + 2fx + 2gy + c = 0\) if axis of ellipse parallel t...

The eccentricity of the ellipseax2+by2+2fx+2gy+c=0ax^{2} + by^{2} + 2fx + 2gy + c = 0 if axis of ellipse parallel to x-axis is

A

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

B

a+bb\sqrt{\frac{a + b}{b}}

C

a+ba\sqrt{\frac{a + b}{a}}

D

None

Answer

None

Explanation

Solution

None of these

ax2+by2+2fx+2gy+c=0ax^{2} + by^{2} + 2fx + 2gy + c = 0

a{x2+2fxa}+b{y2+2gyb}+c=0a\left\{ x^{2} + \frac{2fx}{a} \right\} + b\left\{ y^{2} + \frac{2gy}{b} \right\} + c = 0⇒a(x+fa)2+b(y+gb)2=(f2a+g2b−c)a\left( x + \frac{f}{a} \right)^{2} + b\left( y + \frac{g}{b} \right)^{2} = \left( \frac{f^{2}}{a} + \frac{g^{2}}{b} - c \right)⇒(x+fa)2(f2a+g2b−c)a+(y+gb)2(f2a+g2b−c)b=1\frac{\left( x + \frac{f}{a} \right)^{2}}{\frac{\left( \frac{f^{2}}{a} + \frac{g^{2}}{b} - c \right)}{a}} + \frac{\left( y + \frac{g}{b} \right)^{2}}{\frac{\left( \frac{f^{2}}{a} + \frac{g^{2}}{b} - c \right)}{b}} = 1

if e eccentricity then

⇒ f2a+g2b−cb=f2a+g2b−ca(1−e2)\frac{\frac{f^{2}}{a} + \frac{g^{2}}{b} - c}{b} = \frac{\frac{f^{2}}{a} + \frac{g^{2}}{b} - c}{a}\left( 1 - e^{2} \right)⇒ 1−e2=ab1 - e^{2} = \frac{a}{b} ⇒ e2=b−abe^{2} = \frac{b - a}{b} ∴e=b−ab\therefore e = \sqrt{\frac{b - a}{b}}