The equation of the normal drawn at the point | vedclass

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(C) The given equation of the ellipse is $x^2+2y^2-2x+8y+5=0$. Completing the square,we get $(x^2-2x+1) + 2(y^2+4y+4) = -5+1+8$,which simplifies to $(

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$\sqrt{2}x-y=3+\sqrt{2}$

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(C) The given equation of the ellipse is $x^2+2y^2-2x+8y+5=0$. Completing the square,we get $(x^2-2x+1) + 2(y^2+4y+4) = -5+1+8$,which simplifies to $(x-1)^2 + 2(y+2)^2 = 4$. Dividing by $4$,we have $\frac{(x-1)^2}{4} + \frac{(y+2)^2}{2} = 1$. Here,$a^2=4$ and $b^2=2$. The point is $P(\sqrt{2}+1, -1)$. The slope of the tangent at $(x_1, y_1)$ is given by differentiating the equation: $2x-2 + 4y y' + 8y' = 0$,so $y'(2x-2 + 4y + 8) = 0$ is incorrect; rather $2(x-1) + 4(y+2)y' = 0$,so $y' = -\frac{x-1}{2(y+2)}$. At $P(\sqrt{2}+1, -1)$,$y' = -\frac{\sqrt{2}+1-1}{2(-1+2)} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$. The slope of the normal is $m_n = -\frac{1}{y'} = \sqrt{2}$. The equation of the normal is $y - (-1) = \sqrt{2}(x - (\sqrt{2}+1))$. $y+1 = \sqrt{2}x - 2 - \sqrt{2}$. $\sqrt{2}x - y = 3 + \sqrt{2}$.

The equation of the normal drawn at the point $(\sqrt{2}+1, -1)$ to the ellipse $x^2+2y^2-2x+8y+5=0$ is

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