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  • The foci of the ellipse   <img alt="\mathrm{25(x+1)^2+9(y+2)^2=225}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B25%28x&plus;1%29%5E2&plus;9%28y&plus;2%29

The foci of the ellipse   \mathrm{25(x+1)^2+9(y+2)^2=225}   are at 

Option: 1

(-1, 2) and (-1, -6)


Option: 2

(-2, 1) and (-2, 6)


Option: 3

(-1, -2) and (-2, -1)


Option: 4

(-1, -2) and (-1, -6).


Answers (1)

best_answer

The given equation is   \mathrm{25(x+1)^2+9(y+2)^2=225}
Or,    \mathrm{\quad \frac{(x+1)^2}{9}+\frac{(y+2)^2}{25}=1}
Centre of the ellipse \mathrm{\equiv(-1,-2)}
and \mathrm{\mathrm{a}=3, \mathrm{~b}=5,~ so ~that ~\mathrm{a}<\mathrm{b} \Rightarrow 3=5 \sqrt{1-\mathrm{e}^2}}
or   \mathrm{\mathrm{e}^2=1-9 / 25=16 / 25 \Rightarrow \mathrm{e}=4 / 5}
Hence foci are \mathrm{(-1,-2-5 \times 4 / 5)} and \mathrm{(-1,-2+5 \times 4 / 5)} 
i.e. foci are (-1,-6) and (-1,2).
Hence (A) is the correct answer.

Posted by

shivangi.bhatnagar

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