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The distances from the foci of\mathrm{P\left(x_1, y_1\right)} of on the ellipse \mathrm{\frac{x^2}{9}+\frac{y^2}{25}=1} 

are    

 

Option: 1

\mathrm{4 \pm \frac{5}{4} y_1}


Option: 2

\mathrm{5 \pm \frac{4}{5} x_1}


Option: 3

\mathrm{5 \pm \frac{4}{5} y_1}


Option: 4

None of these


Answers (1)

best_answer

For the given ellipse \mathrm{b>a} , so the two foci lie on y-axis and their coordinates are \mathrm{(0, \pm b e)}
Where \mathrm{b=5, a=3 \text {. So } e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}}
The focal distances of a point \mathrm{\left(x_1, y_1\right)} on the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1},  Where \mathrm{b^2>a^2} are given by \mathrm{b \pm e y_1}  So, Required distances are \mathrm{b \pm e y_1=5 \pm \frac{4}{5} y_1}

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rishi.raj

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