If the normal at an end of a latus rectum of an ellipse passes through an extremity of the

Question:

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies :

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

  2. $e^{2}+e-1=0$

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

  4. $e^{4}+e^{2}-1=0$


Correct Option:

Solution:

$\frac{a^{2} x}{x_{1}}-\frac{b^{2} y}{y_{1}}=a^{2} e^{2}$

$\frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2}} \cdot a=a^{2} e^{2}$

$\frac{a x}{e}-a y=a^{2} e^{2} \Rightarrow \frac{x}{e}-y=a e^{2}$

passes through $(0, \mathrm{~b})$

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

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

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