Question

The eccentricity of the ellipse $25{\mathrm{x}}^2+16{\mathrm{y}}^2-150\mathrm{x}-175=0$ is $?$

• A. $\frac{2}{5}$
• B. $\frac{2}{3}$
• C. $\frac{4}{5}$
• D. $\frac{3}{4}$
• E. $\frac{3}{5}$

Answer 1

The correct option is E

$\frac{3}{5}$

Finding the eccentricity of ellipse:

Given the equation of the ellipse $25{\mathrm{x}}^2+16{\mathrm{y}}^2-150\mathrm{x}-175=0$

Dividing by $25\times 16$ on both sides,

Equation becomes,

$\begin{array}{rcl}& \Rightarrow & \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{25}-\frac{6}{16}\mathrm{x}-\frac{7}{16}=0\\ & \Rightarrow & \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{25}-\frac{6}{16}\mathrm{x}-\frac{16}{16}+\frac{9}{16}=0\\ & \Rightarrow & \frac{{\left[x-3\right]}^2}{16}+\frac{y^2}{25}=1\\ \therefore e& =& \sqrt{1-\frac{a^2}{b^2}}\\ & =& \sqrt{1-\frac{16}{25}}\\ & =& \sqrt{\frac{9}{25}}\\ & =& \frac{3}{5}\end{array}$

Hence, option (E) is the correct answer.