The eccentricity of ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

The eccentricity of ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is
  1. $\frac{7}{16}$
  2. $\frac{5}{4}$
  3. $\frac{\sqrt{7}}{4}$
  4. $\frac{\sqrt{7}}{2}$

Solution

Given ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$ Here, $a^2=16, b^2=9$ Now, $b^2=a^2\left(1-e^2\right)$ $\begin{array}{rlrl} & 9 & =16\left(1-e^2\right) \\ \Rightarrow & & 1-e^2 & =\frac{9}{16} \\ \Rightarrow & e^2 & =1-\frac{9}{16}=\frac{7}{16} \\ \Rightarrow & e & =\frac{\sqrt{7}}{4}\end{array}$

Asked in: AP EAMCET 2001

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