Let e be the eccentricity of the ellipse frac | vedclass

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(C) Given the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$,we have $a=5$ and $b=4$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \fra

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$\frac{6}{e}$

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(C) Given the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$,we have $a=5$ and $b=4$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. The end of the latus rectum in the first quadrant is $(ae, \frac{b^2}{a}) = (5 	imes \frac{3}{5}, \frac{16}{5}) = (3, 3.2)$. The equation of the normal at $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$. Substituting $x_1 = 3, y_1 = 3.2, a^2 = 25, b^2 = 16$: $\frac{25x}{3} - \frac{16y}{3.2} = 25 - 16 = 9$. $\frac{25x}{3} - 5y = 9$. Multiplying by $3$,we get $25x - 15y = 27$. Comparing this with $lx + my = 27$,we get $l = 25$ and $m = -15$. Thus,$l + m = 25 - 15 = 10$. Since $e = 0.6$,we have $\frac{6}{e} = \frac{6}{0.6} = 10$. Therefore,$l + m = \frac{6}{e}$.

Let $e$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. If $a=5, b=4$ and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is $lx+my=27$,then $l+m=$

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