Find the eccentricity of the following ellipse. Round your answer to two decimals.
$$4 x ^ { 2 } + 9 y ^ { 2 } - 24 x + 18 y - 36 = 0$$
A) $1.25$
B) $1.50$
C) $329.06$
D) $0.75$
E) $ 0.67$

Question

Quizplus · Accepted Answer

The given equation is $4x^2 + 9y^2 - 24x + 18y - 36 = 0$. To find the eccentricity of the ellipse, we first complete the squares for $x$ and $y$. Rearranging and grouping the $x$ and $y$ terms gives us: $4(x^2 - 6x) + 9(y^2 + 2y) = 36$.Completing the square for $x^2 - 6x$, we add and subtract $(6/2)^2 = 9$, inside the parenthesis, which gives us $4[(x - 3)^2 - 9]$.For $y^2 + 2y$, we add and subtract $(2/2)^2 = 1$, resulting in $9[(y + 1)^2 - 1]$.Substituting these into the equation, we get $4(x - 3)^2 - 36 + 9(y + 1)^2 - 9 = 36$, or $4(x - 3)^2 + 9(y + 1)^2 = 36 + 36 + 9$, which simplifies to $4(x - 3)^2 + 9(y + 1)^2 = 81$.Dividing the entire equation by 81 to get it in standard form, we have $\frac{(x - 3)^2}{20.25} + \frac{(y + 1)^2}{9} = 1$.The standard form of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2$ is the larger of the two denominators. Here, $a^2 = 20.25$ and $b^2 = 9$.The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$. Substituting $a^2 = 20.25$ and $b^2 = 9$ gives us $e = \sqrt{1 - \frac{9}{20.25}} = \sqrt{1 - \frac{9}{20.25}} = \sqrt{1 - 0.4444} = \sqrt{0.5556} \approx 0.75$.Therefore, the eccentricity of the ellipse is approximately $0.75$.

Find the Eccentricity of the Following Ellipse 4x2+9y2−24x+18y−36=04 x ^ { 2 } + 9 y ^ { 2 } - 24 x + 18 y - 36 = 04x2+9y2−24x+18y−36=0

Find the Eccentricity of the Following Ellipse $4 x ^ { 2 } + 9 y ^ { 2 } - 24 x + 18 y - 36 = 0$