Let the eccentricity of the ellipse 2x^2 + ay | vedclass

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(D) The given equation is $2x^2 + ay^2 - 8x - 2ay + 8 - a = 0$.Rearranging the terms: $2(x^2 - 4x) + a(y^2 - 2y) = a - 8$.Completing the square: $2(x-

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$x - y - 1 \pm \sqrt{\frac{10}{3}} = 0$

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(D) The given equation is $2x^2 + ay^2 - 8x - 2ay + 8 - a = 0$.Rearranging the terms: $2(x^2 - 4x) + a(y^2 - 2y) = a - 8$.Completing the square: $2(x-2)^2 - 8 + a(y-1)^2 - a = a - 8$.$2(x-2)^2 + a(y-1)^2 = 2a$.Dividing by $2a$: $\frac{(x-2)^2}{a} + \frac{(y-1)^2}{2} = 1$.Since the major axis is parallel to the $Y$-axis,the denominator of the $y$-term must be greater than the $x$-term,so $2 > a$.The eccentricity $e = \frac{1}{\sqrt{3}}$ is given by $a = 2(1 - e^2) = 2(1 - \frac{1}{3}) = 2(\frac{2}{3}) = \frac{4}{3}$.The ellipse equation is $\frac{(x-2)^2}{4/3} + \frac{(y-1)^2}{2} = 1$.The equation of a tangent with slope $m=1$ is $y - 1 = m(x - 2) \pm \sqrt{a^2m^2 + b^2}$.Here $A^2 = 4/3$ and $B^2 = 2$. Thus,$y - 1 = 1(x - 2) \pm \sqrt{\frac{4}{3}(1)^2 + 2}$.$y - 1 = x - 2 \pm \sqrt{\frac{4+6}{3}} \Rightarrow y - 1 = x - 2 \pm \sqrt{\frac{10}{3}}$.$x - y - 1 \pm \sqrt{\frac{10}{3}} = 0$.

Let the eccentricity of the ellipse $2x^2 + ay^2 - 8x - 2ay + (8 - a) = 0$ be $\frac{1}{\sqrt{3}}$. If the major axis of this ellipse is parallel to the $Y$-axis,then the equation of the tangent to this ellipse with slope $1$ is:

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