Find the eccentricity $ e $ of the ellipse.
-$$x ^ { 2 } + 2 y ^ { 2 } = 6$$
A) $\sqrt { 2 }$
B) $\sqrt { 6 }$
C) $\frac { \sqrt { 6 } } { 2 }$
D) $\frac { \sqrt { 2 } } { 2 }$

Question

Quizplus · Accepted Answer

The equation of the ellipse is given in the form $Ax^2 + By^2 = C$, where $A$ and $B$ are the coefficients of $x^2$ and $y^2$, respectively, and $C$ is a constant. To find the eccentricity $e$ of the ellipse, we first need to bring the equation to its standard form, which is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2$ and $b^2$ are the denominators of the $x^2$ and $y^2$ terms, respectively, representing the squares of the semi-major and semi-minor axes. For the given equation $x^2 + 2y^2 = 6$, dividing through by 6 gives $\frac{x^2}{6} + \frac{y^2}{3} = 1$, which means $a^2 = 6$ and $b^2 = 3$. The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.Substituting $a^2 = 6$ and $b^2 = 3$ into the formula gives $e = \sqrt{1 - \frac{3}{6}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$.

Find the Eccentricity e e e Of the Ellipse x2+2y2=6x ^ { 2 } + 2 y ^ { 2 } = 6x2+2y2=6

Find the Eccentricity $e$ Of the Ellipse $x ^ { 2 } + 2 y ^ { 2 } = 6$