Question

The eccentricity of ellipse represented by the equation $25x^2+16y^2-150x-175=0$ is :

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

Answer 1

The correct option is B $\frac{3}{4}$

$\begin{array}{ccc}Givenequationofelipse& & \\ 25x^2+16y^2-150x-175=0& & \\ solveterms25x^2-150x+225-225+16y^2-175=0& & \\ let5x-15=x^2& & 4y=y\\ x^2+y^2={\left(20\right)}^2& & \\ \frac{{\left(x\right)}^2}{{\left(20\right)}^2}+\frac{{\left(y\right)}^2}{{\left(20\right)}^2}=1& & \\ s\tan dardformofelipse1mu1mu\frac{x^2}{a^2}+\frac{y^2}{b^2}=1& & \\ now\frac{{\left\{s\left(x-3\right)\right\}}^2}{{\left(20\right)}^2}+\frac{{\left(4y\right)}^2}{{\left(20\right)}^2}=1& & \\ \Rightarrow \frac{{\left(x-3\right)}^2}{{\left(4\right)}^2}+\frac{{\left(y\right)}^2}{{\left(20\right)}^2}=1\left[\begin{array}{c}a=4\\ b=5\end{array}\right]& & \\ now{c}^2=b^2-a^2={\left(5\right)}^2-{\left(4\right)}^2=25-16=9={\left(3\right)}^2& c^2=3^2& \\ c=3& & \\ nowfindecentricity\left(e\right)=\frac{c}{a}=\frac{3}{4}& & \\ nowecentricity\left(e\right)=\frac{3}{4}=0.75& & \end{array}$