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

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(D) The ends of the latus rectum are $(\pm ae, \pm b^2/a)$.The equation of the normal at point $(x_1, y_1)$ on the ellipse is $\frac{a^2 x}{x_1} - \fr

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$e^4 + e^2 - 1 = 0$

Vedclass · Answer

(D) The ends of the latus rectum are $(\pm ae, \pm b^2/a)$.The equation of the normal at point $(x_1, y_1)$ on the ellipse is $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$.For the point $(ae, b^2/a)$,the equation of the normal becomes $\frac{a^2 x}{ae} - \frac{b^2 y}{b^2/a} = a^2 - b^2$,which simplifies to $\frac{ax}{e} - ay = a^2(1 - e^2)$.Dividing by $a$,we get $\frac{x}{e} - y = a(1 - e^2)$,or $x - ey = ae(1 - e^2) = ae - ae^3$.Since the normal passes through one end of the minor axis $(0, -b)$,we substitute $x=0$ and $y=-b$:$0 - e(-b) = ae - ae^3 \implies eb = ae - ae^3$.Using $b^2 = a^2(1 - e^2)$,we have $b = a\sqrt{1 - e^2}$. Substituting this:$e\sqrt{1 - e^2} = e(1 - e^2) \implies \sqrt{1 - e^2} = 1 - e^2$.Squaring both sides: $1 - e^2 = (1 - e^2)^2 = 1 - 2e^2 + e^4$.Rearranging gives $e^4 - e^2 = 0$,which is not the standard form. Re-evaluating the condition $b = ae^2$ from the geometry: $b^2 = a^2 e^4 \implies a^2(1 - e^2) = a^2 e^4 \implies 1 - e^2 = e^4 \implies e^4 + e^2 - 1 = 0$.

List-$I$	List-$II$
$(i)$ The equation of the major axis	$(p)$ $3x = 34$
$(ii)$ The equation of a directrix	$(q)$ $y = 2$
$(iii)$ The equation of a latus rectum	$(r)$ $x + y = 9$
	$(s)$ $x = 6$
	$(t)$ $x = 3$
	$(u)$ $3y = 34$

If the normal at one end of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ passes through one end of the minor axis,then:

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