10-2. Parabola, Ellipse, Hyperbola
hard

જો ઉપવલયના નાભીલંબની લંબાઈ $4\,એકમ$ અને નાભી અને મુખ્યઅક્ષ પરના નજીકના શિરોબિંદુ વચ્ચેનું અંતર $\frac {3}{2}\,એકમ$ હોય તો ઉત્કેન્દ્ર્તા મેળવો. 

A

$\frac {1}{2}$

B

$\frac {2}{3}$

C

$\frac {1}{9}$

D

$\frac {1}{3}$

(JEE MAIN-2018)

Solution

Let for ellipse coordinates of focus and vertex are $(ae,0)$ and $(a,0)$ respectively.

$\therefore $ Distance between focus and vertex

$ = a\left( {1 – e} \right) = \frac{3}{2}$                       (given)

$ \Rightarrow a – \frac{3}{2} = ae$

$ \Rightarrow {a^2} + \frac{9}{4} – 3a = {a^2}{e^2}\,\,\,\,\,\,\,……\left( i \right)$

Length of latus rectum $ = \frac{{2{b^2}}}{a} = 4$

$ \Rightarrow {b^2} = 2a\,\,\,\,\,\,\,\,\,\,……\left( {ii} \right)$

         ${e^2} = 1 – \frac{{{b^2}}}{{{a^2}}}$

$ \Rightarrow {e^2} = 1 – \frac{{2a}}{{{a^2}}}$         (from$(ii)$)

$ \Rightarrow {e^2} = 1 – \frac{2}{a}\,\,\,\,\,\,\,\,\,…….\left( 3 \right)$

Substituting the value of ${e^2}$ in eq. $(i)$ we get;

$ \Rightarrow {a_2} + \frac{9}{4} – 3a = {a^2}\left( {1 – \frac{2}{a}} \right)$

$ \Rightarrow a = \frac{9}{4}$

$\therefore $ from eq. $(iii)$ we get;

${e^2} = 1 – \frac{2}{a} = 1 – \frac{8}{9} = \frac{1}{9}$

$ \Rightarrow e = \frac{1}{3}$

Standard 11
Mathematics

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