Question

The equation $3(x+y-5{)}^2+2(x-y+7{)}^2=6$ represents an ellipse then eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

$\frac{(x+y-5{)}^2}{2}+\frac{(x-y+7{)}^2}{3}=1\frac{{\left[\frac{x+y-5}{\sqrt{2}}\right]}^2}{1}+\frac{{\left[\frac{x-y+7}{\sqrt{2}}\right]}^2}{\frac{3}{2}}=1b^2=1;a^2=3/2b^2=a^2(1-e^2)1=\frac{3}{2}(1-e^2)\Rightarrow e=\frac{1}{\sqrt{3}}$