An ellipse frac{x^2}{a^2}+frac{y^2}{b^2}=1 wi | vedclass

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(D) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Given eccentricity $e=\frac{2 \sqrt{2}}{3}$. The circle is $x^2+y^2=18$,so its

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$\frac{x^2}{18}+\frac{9y^2}{2}=1$

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(D) The equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Given eccentricity $e=\frac{2 \sqrt{2}}{3}$. The circle is $x^2+y^2=18$,so its radius $R=\sqrt{18}=3 \sqrt{2}$ and diameter $D=6 \sqrt{2}$. The major axis length $2a=6 \sqrt{2}$,so $a=3 \sqrt{2}$ and $a^2=18$. Using $e^2=1-\frac{b^2}{a^2}$,we have $\frac{8}{9}=1-\frac{b^2}{18}$,which gives $\frac{b^2}{18}=\frac{1}{9}$,so $b^2=2$. The ellipse is $\frac{x^2}{18}+\frac{y^2}{2}=1$. Let $(h, k)$ be the pole of a tangent to the circle. The polar line with respect to the ellipse is $\frac{xh}{18}+\frac{yk}{2}=1$. This line is a tangent to the circle $x^2+y^2=18$. The condition for the line $lx+my=1$ to be a tangent to $x^2+y^2=R^2$ is $R^2(l^2+m^2)=1$. Here $l=\frac{h}{18}$ and $m=\frac{k}{2}$,so $18(\frac{h^2}{18^2}+\frac{k^2}{4})=1$. Simplifying,$\frac{h^2}{18}+\frac{18k^2}{4}=1$,which is $\frac{h^2}{18}+\frac{9k^2}{2}=1$. Thus,the locus is $\frac{x^2}{18}+\frac{9y^2}{2}=1$.

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e=\frac{2 \sqrt{2}}{3}$ is inscribed in a circle $x^2+y^2=18$ such that the length of its major axis is equal to the diameter of this circle. The locus of the poles of all the tangents of the circle with respect to the ellipse is

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