For k0,the shortest distance from a point P( | vedclass

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(C) Given the ellipse equation: $9x^2+4y^2-18x+16y-11=0$. Since $P(1, k)$ lies on the ellipse,substitute $x=1$: $9(1)^2+4k^2-18(1)+16k-11=0$ $9+4k^2-1

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$\frac{9}{\sqrt{5}}-3$

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(C) Given the ellipse equation: $9x^2+4y^2-18x+16y-11=0$. Since $P(1, k)$ lies on the ellipse,substitute $x=1$: $9(1)^2+4k^2-18(1)+16k-11=0$ $9+4k^2-18+16k-11=0$ $4k^2+16k-20=0$ $k^2+4k-5=0$ $(k+5)(k-1)=0$. Since $k>0$,we have $k=1$. Thus,$P$ is $(1, 1)$. Rewrite the ellipse equation in standard form: $9(x-1)^2+4(y+2)^2 = 11+9+16 = 36$ $\frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$. Here $a^2=4$ and $b^2=9$,so $b>a$. Eccentricity $e = \sqrt{1-\frac{a^2}{b^2}} = \sqrt{1-\frac{4}{9}} = \frac{\sqrt{5}}{3}$. The directrices are $y+2 = \pm \frac{b}{e} = \pm \frac{3}{\sqrt{5}/3} = \pm \frac{9}{\sqrt{5}}$. So,$y = -2 \pm \frac{9}{\sqrt{5}}$. The distances from $P(1, 1)$ to the directrices $y = -2 + \frac{9}{\sqrt{5}}$ and $y = -2 - \frac{9}{\sqrt{5}}$ are: $d_1 = |1 - (-2 + \frac{9}{\sqrt{5}})| = |3 - \frac{9}{\sqrt{5}}| = \frac{9}{\sqrt{5}} - 3$ (since $\frac{9}{\sqrt{5}} \approx 4.02 > 3$). $d_2 = |1 - (-2 - \frac{9}{\sqrt{5}})| = |3 + \frac{9}{\sqrt{5}}| = 3 + \frac{9}{\sqrt{5}}$. The shortest distance is $\frac{9}{\sqrt{5}} - 3$.

For $k>0$,the shortest distance from a point $P(1, k)$ on the ellipse $9x^2+4y^2-18x+16y-11=0$ to one of its directrices is

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