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  • The eccentricity of the ellipse which meets the straight line <img alt="\mathrm{\frac{x}{7}+\frac{y}{2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

The eccentricity of the ellipse which meets the straight line \mathrm{\frac{x}{7}+\frac{y}{2}=1} on the axis of x and the straight line\mathrm{\frac{x}{3}-\frac{y}{5}=1} on the axis of y and whose axes lie among the axes of co-ordinates is 

 

Option: 1

\frac{3 \sqrt{2}}{7}


Option: 2

\frac{2 \sqrt{6}}{7}


Option: 3

\frac{ \sqrt{6}}{7}


Option: 4

none of these

 


Answers (1)

best_answer

Let the equation of the ellipse be \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} .

It is given that it passes through (7, 0) and (0, –5)

\mathrm{\therefore \quad a^2=49 \text { and } b^2=25}

\mathrm{\text { Since } b^2=a^2\left(1-e^2\right)}

\mathrm{\begin{gathered} \therefore \quad 25=49\left(1-e^2\right) \\ \Rightarrow \quad 1-e^2=\frac{25}{49} \Rightarrow \quad e^2=1-\frac{25}{49}=\frac{24}{49} \\ \Rightarrow e=\frac{2 \sqrt{6}}{7} \end{gathered}}

 

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Nehul

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