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  • If the normal at the end of a latus rectum of an ellipse  <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}}" src="https://entrancecorner.oncodecogs.com/gif.lat

If the normal at the end of a latus rectum of an ellipse  \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}}  = 1 passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by the equation

 

Option: 1

\mathrm{e^2+e-1=0}


Option: 2

\mathrm{e^2+e+1=0}


Option: 3

\mathrm{e^4+e^2+1=0}


Option: 4

\mathrm{e^4+e^2-1=0}


Answers (1)

best_answer

\mathrm{ \text { Normal at }\left(a e, \frac{b^2}{a}\right) \text { is } \frac{a x}{e}-a y=a^2-b^2 \\ }

\mathrm{ \text { It passes through }(0,-b) \\ }

\mathrm{ \Rightarrow a b=a^2-b^2=a^2 e^2 \Rightarrow b=a e^2 \Rightarrow b^2=a^2 e^4 \\ }

\mathrm{ \Rightarrow a^2\left(1-e^2\right)=a^2 e^4 \Rightarrow e^4+e^2-1=0 . }

Hence (D) is the correct answer.

Posted by

chirag

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