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

Question

Quizplus · Accepted Answer

First we need to write the equation in standard form by completing the square:
\begin{align*}
2x^2-4x+5y^2+10y-10 &= 0 \
2(x^2-2x)+5(y^2+2y) &= 10 \
2(x^2-2x+1)+5(y^2+2y+1) &= 10+2+5 \
2(x-1)^2+5(y+1)^2 &= 17
\end{align*}
Thus, $\frac{(x-1)^2}{\frac{17}{2}}+\frac{(y+1)^2}{\frac{17}{5}} = 1$, and we can see that $a^2=\frac{17}{2}$ and $b^2 = \frac{17}{5}$. Therefore, $c=\sqrt{a^2-b^2}=\sqrt{\frac{59}{10}}$, and the eccentricity is $e = \frac{c}{a} = \frac{\sqrt{\frac{59}{10}}}{\sqrt{\frac{17}{2}}} \approx 0.77$.

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 $2 x ^ { 2 } + 5 y ^ { 2 } - 4 x + 10 y - 10 = 0$