Statement I: The equation of the directrix of | vedclass

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(A) For Statement $I$: The equation $4x^2+y^2-8x-4y+4=0$ can be written as $4(x-1)^2-4 + (y-2)^2-4+4=0$,which simplifies to $4(x-1)^2+(y-2)^2=4$,or $\

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Statement $I$ is true,but Statement $II$ is false

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(A) For Statement $I$: The equation $4x^2+y^2-8x-4y+4=0$ can be written as $4(x-1)^2-4 + (y-2)^2-4+4=0$,which simplifies to $4(x-1)^2+(y-2)^2=4$,or $\frac{(x-1)^2}{1} + \frac{(y-2)^2}{4} = 1$.Here $a^2=1, b^2=4$,so $b > a$. The eccentricity $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.The directrices are $y-k = \pm \frac{b}{e}$,so $y-2 = \pm \frac{2}{\sqrt{3}/2} = \pm \frac{4}{\sqrt{3}}$.Thus $y = 2 \pm \frac{4}{\sqrt{3}}$,which gives $y = \frac{6 \pm 4\sqrt{3}}{3}$,or $3y = 6 \pm 4\sqrt{3}$. Statement $I$ is true.For Statement $II$: The equation $x^2+4y^2-4x-8y+4=0$ can be written as $(x-2)^2-4 + 4(y-1)^2-4+4=0$,which simplifies to $(x-2)^2+4(y-1)^2=4$,or $\frac{(x-2)^2}{4} + \frac{(y-1)^2}{1} = 1$.Here $a^2=4, b^2=1$,so $a > b$. The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$.The latus rectum is $x-h = \pm ae$,so $x-2 = \pm 2(\frac{\sqrt{3}}{2}) = \pm \sqrt{3}$.This gives $x = 2 \pm \sqrt{3}$. Statement $II$ claims the latus rectum is $y=2+\sqrt{3}$,which is incorrect. Statement $II$ is false.

Statement $I$: The equation of the directrix of the ellipse $4x^2+y^2-8x-4y+4=0$ is $3y=6-4\sqrt{3}$.
Statement $II$: The equation of the latus rectum of the ellipse $x^2+4y^2-4x-8y+4=0$ is $y=2+\sqrt{3}$.
Which of the above statement$(s)$ is (are) true?

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Statement $I$: The equation of the directrix of the ellipse $4x^2+y^2-8x-4y+4=0$ is $3y=6-4\sqrt{3}$.Statement $II$: The equation of the latus rectum of the ellipse $x^2+4y^2-4x-8y+4=0$ is $y=2+\sqrt{3}$.Which of the above statement$(s)$ is (are) true?