A) \[24\]
B) \[32\]
C) \[64\]
D) \[48\]
Correct Answer: C
Solution :
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 Eq. (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 \[=2(a+b)=2(20+12)=64\]
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