What is the equation of the ellipse whose vertices are $( \pm 5,0)$ and foci are at $( \pm 4,0)$ ?

• Correct answer
  A

  $\frac{x^2}{25}+\frac{y^2}{9}=1$
• B

  $\frac{x^2}{16}+\frac{y^2}{9}=1$
• C

  $\frac{x^2}{25}+\frac{y^2}{16}=1$
• D

  $\frac{x^2}{9}+\frac{y^2}{25}=1$

Solution

Given, vertices, $(5,0)$ and $(-5,0)$ foci $(4,0)$ and $(-4,0)$
i.e., $a \mathrm{e}=4$ $\Rightarrow \mathrm{e}=\frac{4}{\mathrm{a}}=\frac{4}{5}$ we know, $\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)$ $\begin{aligned} & =5^2\left(1-\left(\frac{4}{5}\right)^2\right) \\ & =25\left(\frac{25-16}{25}\right)=9 \\ & \therefore \mathrm{a}^2=25, \mathrm{~b}^2=9 \end{aligned}$ Equation of ellipse is $\frac{x^2}{25}+\frac{y^2}{9}=1$

Hence, the correct answer is option: 'A'.