Find the eccentricity of the following ellipse. Round your answer to two decimals.
$$2 x ^ { 2 } + 5 y ^ { 2 } - 4 x + 10 y - 10 = 0$$
A) $1.50$
B) $1.07$
C) $60.69$
D) $0.77$
E) $ 0.63$

Question

Quizplus · Accepted Answer

The given equation is $2x^2 + 5y^2 - 4x + 10y - 10 = 0$. To find the eccentricity of the ellipse, we first need to complete the square for both $x$ and $y$ terms and bring the equation to its standard form.Rearranging and grouping $x$ and $y$ terms:$$2(x^2 - 2x) + 5(y^2 + 2y) = 10$$Completing the square:$$2[(x - 1)^2 - 1] + 5[(y + 1)^2 - 1] = 10$$$$2(x - 1)^2 + 5(y + 1)^2 = 10 + 2 + 5$$$$2(x - 1)^2 + 5(y + 1)^2 = 17$$Dividing through by 17 to get the standard form of an ellipse:$$\frac{(x - 1)^2}{\frac{17}{2}} + \frac{(y + 1)^2}{\frac{17}{5}} = 1$$So, the semi-major axis $a^2 = \frac{17}{2}$ and the semi-minor axis $b^2 = \frac{17}{5}$. The eccentricity $e$ of an ellipse is given by $e = \sqrt{1 - \frac{b^2}{a^2}}$.Substituting $a^2$ and $b^2$:$$e = \sqrt{1 - \frac{\frac{17}{5}}{\frac{17}{2}}}$$$$e = \sqrt{1 - \frac{2}{5}}$$$$e = \sqrt{\frac{3}{5}}$$$$e = \sqrt{0.6}$$$$e \approx 0.77$$Therefore, the eccentricity of the ellipse is approximately 0.77.

Find the Eccentricity of the Following Ellipse 2x2+5y2−4x+10y−10=02 x ^ { 2 } + 5 y ^ { 2 } - 4 x + 10 y - 10 = 02x2+5y2−4x+10y−10=0

Find the Eccentricity of the Following Ellipse $2 x ^ { 2 } + 5 y ^ { 2 } - 4 x + 10 y - 10 = 0$