A) \[x+ey+{{e}^{3}}a=0\]
B) \[x-ey-{{e}^{3}}a=0\]
C) \[x+ey-{{e}^{2}}a=0\]
D) None of these
Correct Answer: B
Solution :
The equation of the normal at \[({{x}_{1}},\,{{y}_{1}})\] to the given ellipse is \[\frac{{{a}^{2}}x}{{{x}_{1}}}-\frac{{{b}^{2}}y}{{{y}^{1}}}={{a}^{2}}-{{b}^{2}}\] We know that coordinate of the latus rectum at postive end is \[\left( ae,\,\,\frac{{{b}^{2}}}{a} \right)\] Then, \[{{x}_{1}}=ae\] and \[{{y}_{1}}=\frac{{{b}^{2}}}{a}\] Therefore, the equation of the normal at postive end of the latusrectum is \[\frac{{{a}^{2}}x}{ae}-\frac{{{b}^{2}}y}{{{b}^{2}}/a}={{a}^{2}}{{e}^{2}}\] \[[\because \,{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})]\] \[\Rightarrow \] \[\frac{ax}{e}-ay={{a}^{2}}{{e}^{2}}\] \[\Rightarrow \] \[x-ey-{{e}^{3}}a=0\]
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