Question

The sum of lengths of major and minor axes- of an ellipse whose eccentricity is $\frac{4}{5}$ and length of latuserectum is $14.4$ , is

• A. $24$
• B. $32$
• C. $64$
• D. $48$

Answer 1

Answer: (C)

$64$

Answer 2

Given, eccentricity of an ellipse $e=\frac{4}{5}$
$\Rightarrow$ $\frac{c}{a}=\frac{4}{5}$ $[\because \,\,c=ae]$
$\Rightarrow$ $c=\frac{4a}{5}$ ..(i) and length of latuserectum $\frac{2{{b}^{2}}}{a}=14.4$
$\Rightarrow$ $\frac{{{b}^{2}}}{a}=7.2$
$\Rightarrow$ $\frac{{{b}^{2}}}{a}=\frac{72}{10}=\frac{36}{6}$
$\Rightarrow$ ${{b}^{2}}=\frac{36\,a}{5}$ .. (ii) In an ellipse, we know that ${{c}^{2}}={{a}^{2}}-{{b}^{2}}$
$\Rightarrow$ ${{\left( \frac{4a}{5} \right)}^{2}}={{a}^{2}}-\frac{36a}{5}$
$\Rightarrow$ $\frac{16{{a}^{2}}}{25}={{a}^{2}}-\frac{36a}{5}$ [using Eqs. (i) and (ii)]
$\Rightarrow$ $\frac{16{{a}^{2}}-25{{a}^{2}}}{25}=\frac{-36a}{5}$
$\Rightarrow$ $\frac{-9{{a}^{2}}}{25}=-\frac{36\,a}{5}$
$\Rightarrow$ $\frac{a}{5}=4$
$\Rightarrow$ $a=20$ Then, from E (ii), we get ${{b}^{2}}=\frac{36}{5}\times 20$
$\Rightarrow$ ${{b}^{2}}=144\Rightarrow b=\pm 12$
$\Rightarrow$ $b=12$ $[\because \,\,b\ne -12]$ Hence, sum of major and minor axes
$=2(a+b)=2(20+12)=64$