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यदि $OB$, एक दीर्घवृत्त का अर्ध लघुअक्ष है, $F _{1}$ तथा $F _{2}$ उसकी नाभियाँ हैं तथा $F _{1} B$ तथा $F _{2} B$ के बीच का कोण एक समकोण है, तो दीर्घवृत्त की उत्केंद्रता का वर्ग है
$\frac{1}{2}$
$\frac{1}{{\sqrt 2 }}$
$\frac{1}{{2\sqrt 2 }}$
$\frac{1}{4}$
Solution
Let $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ be he equation of ellipse.
Given that ${F_1}B$ and ${F_2}B$ are perpandicluar to each other.
Slope of ${F_1}B \times $ alop of ${F_2}B = – 1$
$\left( {\frac{{0 – b}}{{ – ae – 0}}} \right) \times \left( {\frac{{0 – b}}{{ae – 0}}} \right) = – 1$
$\left( {\frac{b}{{ae}}} \right) \times \left( {\frac{{ – b}}{{ae}}} \right) = – 1$
${b^2} = {a^2}{e^2}$
${e^2} = \frac{{{b^2}}}{{{a^2}}}\,\,\,$ {$\because $ $\,\,{e^2} = 1 – \frac{{{b^2}}}{{{a^2}}}\,\,$}
$1 – \frac{{{b^2}}}{{{a^2}}}\,\,\, = \frac{{{b^2}}}{{{a^2}}}\,$
$1 = 2\frac{{{b^2}}}{{{a^2}}}\,\,\, \Rightarrow \frac{{{b^2}}}{{{a^2}}}\,\,\, = \frac{1}{2}$
${e^2} = 1 – \frac{{{b^2}}}{{{a^2}}}\,\,\, = 1 – \frac{1}{2} = \frac{1}{2}$
${e^2} = \frac{1}{2}$
