Let $e_{1}$ and $e_{2}$ are the eccentricities of $16 x^{2} + 9 y^{2} = 144$ and $16 x^{2} - 9 y^{2} = 144$ respectively, then which of the following is/are correct

e_{1}^{2} + e_{2}^{2} > 3

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e_{1}^{2} + e_{2}^{2} < 3

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e_{1}^{2} e_{2}^{2} > 1

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\frac{1}{e_{1}^{2}} + \frac{l}{e_{2}^{2}} > 3

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Solution

The correct options are
A $e_{1}^{2} + e_{2}^{2} > 3$
C $e_{1}^{2} e_{2}^{2} > 1$
$16 x^{2} + 9 y^{2} = 144 \Rightarrow \frac{x^{2}}{9} + \frac{y^{2}}{16} = 1$
$a^{2} = 9, b^{2} = 16, b > a$
$e_{1}^{2} = 1 - \frac{a^{2}}{b^{2}} = 1 - \frac{9}{16} = \frac{7}{16}$

$16 x^{2} - 9 y^{2} = 144 \Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$
$a^{2} = 9, b^{2} = 16$
$e_{2}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{16}{9} = \frac{25}{9}$

$e_{1}^{2} + e_{2}^{2} = \frac{7}{16} + \frac{25}{9} = \frac{463}{144} > 3$

$e_{1}^{2} e_{2}^{2} = \frac{7}{16} \times \frac{25}{9} = \frac{175}{144} > 1$

$\frac{1}{e_{1}^{2}} + \frac{1}{e_{2}^{2}} = \frac{16}{7} + \frac{9}{25} = \frac{463}{175} < 3$

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Similar questions

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\frac{x^{2}}{18} + \frac{y^{2}}{4} = 1

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Let e1 and e2 are the eccentricities of 16x2+9y2=144 and 16x2−9y2=144 respectively, then which of the following is/are correct

The correct options are A e21+e22>3 C e21e22>116x2+9y2=144⇒x29+y216=1 a2=9, b2=16, b>a e21=1−a2b2=1−916=716 16x2−9y2=144⇒x29−y216=1 a2=9, b2=16 e22=1+b2a2=1+169=259 e21+e22=716+259=463144>3 e21e22=716×259=175144>1 1e21+1e22=167+925=463175<3

Let $e_{1}$ and $e_{2}$ are the eccentricities of $16 x^{2} + 9 y^{2} = 144$ and $16 x^{2} - 9 y^{2} = 144$ respectively, then which of the following is/are correct