Question

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$- and $e_2$ is the eccentricity of the hyperbola $\frac{x^2}{9}-\frac{y^2}{7}=1$, then $e_1+e_2=$

• A. $\frac{16}{7}$
• B. $\frac{25}{4}$
• C. $\frac{25}{12}$
• D. $\frac{16}{9}$
• E. $\frac{23}{16}$

Answer 1

The correct option is C

$\frac{25}{12}$

The explanation for the correct option:

Step 1. Find the value of $a^2,b^2.$:

The equation of the ellipse

$\begin{array}{rcl}(x^2/a^2)+(y^2/b^2)& =& 1.\\ (x^2/16)+(y^2/7)& =& 1\\ a^2& =& 16,\\ b^2& =& 7\end{array}$

Step 2. Find the Eccentricity ${\mathit{e}}_{\mathbf{1}}$

$\begin{array}{rcl}{e}_1& =& \frac{\sqrt{a^2–b^2}}{\sqrt{a^2}}\\ & =& \frac{\sqrt{16–7}}{\sqrt{16}}\\ & =& \sqrt{\frac{9}{16}}\\ & =& \frac{3}{4}\end{array}$

Step 3. Find the value of $a^2,b^2.$ of hyperbola:

The equation of hyperbola $(x^2/a^2)–(y^2/b^2)=1$

It is given that

$\begin{array}{rcl}(x^2/9)–(y^2/7)& =& 1\\ a^2& =& 9,\\ b^2& =& 7\end{array}$

Step 4. Find the Eccentricity ${\mathit{e}}_{\mathbf{2}}$

$\begin{array}{rcl}{e}_2& =& \sqrt{\frac{a^2+b^2}{a^2}}\\ & =& \sqrt{\frac{9+7}{9}}\\ & =& \frac{4}{3}\\ & & \end{array}$

$\begin{array}{rcl}\therefore e_1+e_2& =& (3/4)+(4/3)\\ & =& \mathbf{25}\mathbf{}\mathbf{/}\mathbf{}\mathbf{12}\end{array}$

Hence, The correct option is (C)