If the perpendicular distance from the focus | vedclass

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(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 9$ $(a = 3)$ and $b < 3$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)

Vedclass · Accepted Answer

$-\frac{2}{3}$

Vedclass · Answer

(A) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 9$ $(a = 3)$ and $b The distance from the focus $(ae, 0)$ to the directrix $x = \frac{a}{e}$ is $\frac{a}{e} - ae = \frac{a(1 - e^2)}{e} = \frac{b^2}{ae}$.Given the distance is $\frac{4}{\sqrt{5}}$,we have $\frac{b^2}{3e} = \frac{4}{\sqrt{5}}$.Substituting $b^2 = 9(1 - e^2)$,we get $\frac{9(1 - e^2)}{3e} = \frac{3(1 - e^2)}{e} = \frac{4}{\sqrt{5}}$.Let $e^2 = t$. Then $3(1 - t) = \frac{4}{\sqrt{5}}\sqrt{t}$. Squaring both sides,$9(1 - t)^2 = \frac{16}{5}t$,which simplifies to $45t^2 - 106t + 45 = 0$.Factoring gives $(9t - 5)(5t - 9) = 0$. Since $b Then $b^2 = 9(1 - \frac{5}{9}) = 4$,so $b = 2$.The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$.The slope of the tangent at $(x_1, y_1)$ is $m = -\frac{b^2 x_1}{a^2 y_1} = -\frac{4(3/\sqrt{2})}{9(2/\sqrt{2})} = -\frac{12}{18} = -\frac{2}{3}$.

If the perpendicular distance from the focus of an ellipse $\frac{x^2}{9} + \frac{y^2}{b^2} = 1$ $(b < 3)$ to its corresponding directrix is $\frac{4}{\sqrt{5}}$,then the slope of the tangent to this ellipse drawn at $\left(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$ is

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