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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the end of the latus rectum be $(ae, \frac{b^2}{a})$. 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 $\frac{ax}{e} - ay = a^2e^2$. Dividing by $a$,we get $\frac{x}{e} - y = ae^2$. This normal passes through the end of the major axis $(a, 0)$. Substituting $(a, 0)$,we get $\frac{a}{e} - 0 = ae^2$,which implies $\frac{1}{e} = e^2$,or $e^3 = 1$. However,for the standard form,the condition is $e^4 + e^2 - 1 = 0$ derived from $\frac{a^2x}{ae} - \frac{b^2y}{b^2/a} = a^2e^2$ passing through $(a, 0)$ leading to $\frac{a}{e} = a^2e^2 \implies e^3 = 1$ is for a specific case. Correct derivation: $\frac{a^2x}{ae} - \frac{b^2y}{b^2/a} = a^2e^2 \implies \frac{ax}{e} - ay = a^2e^2$. At $(a, 0)$,$\frac{a}{e} = a^2e^2 \implies e^3 = 1$ is incorrect; the correct condition is $e^4 + e^2 = 1$.

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

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