The focal distances of the point left(frac{4} | vedclass

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(D) Given the ellipse equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Since $b^2 > a^2$ $(9 > 4)$,this is a vertical ellipse with $a^2 = 4$ and $b^2 = 9

Vedclass · Accepted Answer

$4, 2$

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(D) Given the ellipse equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$. Since $b^2 > a^2$ $(9 > 4)$,this is a vertical ellipse with $a^2 = 4$ and $b^2 = 9$. The eccentricity $e$ is given by $a^2 = b^2(1 - e^2)$,so $4 = 9(1 - e^2)$ $\Rightarrow 1 - e^2 = \frac{4}{9}$ $\Rightarrow e^2 = \frac{5}{9}$ $\Rightarrow e = \frac{\sqrt{5}}{3}$. The foci are $(0, \pm be) = (0, \pm 3 	imes \frac{\sqrt{5}}{3}) = (0, \pm \sqrt{5})$. Let $S = (0, \sqrt{5})$ and $S' = (0, -\sqrt{5})$. Let $P = \left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}ight)$. The focal distances are $PS$ and $PS'$. $PS = \sqrt{\left(\frac{4}{\sqrt{5}} - 0ight)^2 + \left(\frac{3}{\sqrt{5}} - \sqrt{5}ight)^2} = \sqrt{\frac{16}{5} + \left(\frac{3-5}{\sqrt{5}}ight)^2} = \sqrt{\frac{16}{5} + \frac{4}{5}} = \sqrt{\frac{20}{5}} = 2$. $PS' = \sqrt{\left(\frac{4}{\sqrt{5}} - 0ight)^2 + \left(\frac{3}{\sqrt{5}} + \sqrt{5}ight)^2} = \sqrt{\frac{16}{5} + \left(\frac{3+5}{\sqrt{5}}ight)^2} = \sqrt{\frac{16}{5} + \frac{64}{5}} = \sqrt{\frac{80}{5}} = \sqrt{16} = 4$. Thus,the focal distances are $4$ and $2$.

The focal distances of the point $\left(\frac{4}{\sqrt{5}}, \frac{3}{\sqrt{5}}\right)$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ are

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