If the normal at one end of a latus rectum of | vedclass

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(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 32$. One end of the latus rectum is $(ae, \frac{b^2}{a})$. The

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(B) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 32$. One end of the latus rectum is $(ae, \frac{b^2}{a})$. The equation of the normal at $(x_1, y_1)$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $(x_1, y_1) = (ae, \frac{b^2}{a})$,we get $\frac{a^2x}{ae} - \frac{b^2y}{b^2/a} = a^2 - b^2$,which simplifies to $\frac{ax}{e} - ay = a^2 - b^2$. Since this normal passes through the end of the minor axis $(0, b)$,we have $\frac{a(0)}{e} - a(b) = a^2 - b^2$,so $-ab = a^2 - b^2$,or $b^2 - ab - a^2 = 0$. Dividing by $a^2$,we get $(\frac{b}{a})^2 - (\frac{b}{a}) - 1 = 0$. Since $b^2 = a^2(1-e^2)$,we have $(\frac{b}{a})^2 = 1-e^2$. Thus,$(1-e^2) - \sqrt{1-e^2} - 1 = 0$,which gives $\sqrt{1-e^2} = -e^2$. Squaring both sides,$1-e^2 = e^4$,so $1 = e^4 + e^2$. We need to find $\frac{e^4}{1-e^2}$. Since $1-e^2 = e^4$,the expression becomes $\frac{e^4}{e^4} = 1$.

If the normal at one end of a latus rectum of the ellipse $\frac{x^2}{32}+\frac{y^2}{b^2}=1$ passes through one end of the minor axis,then $\frac{e^4}{1-e^2}=$ (Here $e$ is the eccentricity of the ellipse)

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