If the normal drawn at one end of the latus r | vedclass

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(C) The equation of the ellipse is $b^2 x^2 + a^2 y^2 = a^2 b^2$,which can be written as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.One end of the latus

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$e^4 + e^2 = 1$

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(C) The equation of the ellipse is $b^2 x^2 + a^2 y^2 = a^2 b^2$,which can be written as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.One end of the latus rectum is $(ae, \frac{b^2}{a})$.The equation of the normal at $(x_1, y_1)$ is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$.Substituting $(ae, \frac{b^2}{a})$ into the normal equation:$\frac{a^2 x}{ae} - \frac{b^2 y}{b^2/a} = a^2 - b^2 \Rightarrow \frac{ax}{e} - ay = a^2 - b^2$.Since the normal passes through one end of the minor axis $(0, -b)$,we substitute $x=0$ and $y=-b$:$0 - a(-b) = a^2 - b^2 \Rightarrow ab = a^2 - b^2$.Using $b^2 = a^2(1 - e^2)$,we have $b^2 = a^2 - a^2 e^2$,so $a^2 - b^2 = a^2 e^2$.Thus,$ab = a^2 e^2 \Rightarrow b = ae^2$.Squaring both sides,$b^2 = a^2 e^4$.Since $b^2 = a^2(1 - e^2)$,we get $a^2(1 - e^2) = a^2 e^4$.Dividing by $a^2$,$1 - e^2 = e^4 \Rightarrow e^4 + e^2 = 1$.

If the normal drawn at one end of the latus rectum of the ellipse $b^2 x^2 + a^2 y^2 = a^2 b^2$ with eccentricity $e$ passes through one end of the minor axis,then:

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