Eccentricity of ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\mathbf{1}\), if it passes through point (9, 5) and (12, 4) is

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  1. \(\sqrt{3 / 4} \)
  2. \(\sqrt{4 / 5}\)
  3. \(\sqrt{5 / 6}\)
  4. \(\sqrt{6 / 7}\)

Answer (Detailed Solution Below)

Option 4 : \(\sqrt{6 / 7}\)
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Detailed Solution

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Calculation 

Given \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\mathbf{1}\)

At (9, 5)

\(\frac{81}{a^2} + \frac{25}{b^2} = 1\) ......(1)

At (12,4)

\(\frac{144}{a^2} + \frac{16}{b^2} = 1\) .......(2)

From eq. (2) - eq. (1):

\(\frac{63}{a^2} - \frac{9}{b^2} = 0\)

⇒ \(\frac{b^2}{a^2} = \frac{1}{7}\)

⇒ \(\frac{b}{a} = \frac{1}{\sqrt{7}}\)

e = \(\sqrt{1 - \frac{b^2}{a^2}}\) = \(\sqrt{1 - \frac{1}{7}}\) = \(\sqrt{\frac{6}{7}}\)

Hence option 4 is correct

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