If x+sqrt{3} y=3 is the tangent to the ellips | vedclass

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(D) Given the tangent line $x+\sqrt{3} y=3$ ... $(i)$ and the ellipse $2 x^2+3 y^2=k$. The equation of the tangent at point $P(x_1, y_1)$ is $2 x x_1+

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$3 x-\sqrt{3} y=1$

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(D) Given the tangent line $x+\sqrt{3} y=3$ ... $(i)$ and the ellipse $2 x^2+3 y^2=k$. The equation of the tangent at point $P(x_1, y_1)$ is $2 x x_1+3 y y_1=k$ ... (ii). Comparing $(i)$ and (ii),we have $\frac{2 x_1}{1} = \frac{3 y_1}{\sqrt{3}} = \frac{k}{3}$. Thus,$x_1 = \frac{k}{6}$ and $y_1 = \frac{k}{3 \sqrt{3}}$. Since $P(x_1, y_1)$ lies on the ellipse,$2(\frac{k}{6})^2 + 3(\frac{k}{3 \sqrt{3}})^2 = k$. $\frac{k^2}{18} + \frac{k^2}{9} = k$ $\Rightarrow \frac{k^2}{6} = k$ $\Rightarrow k = 6$. So,$x_1 = 1$ and $y_1 = \frac{6}{3 \sqrt{3}} = \frac{2}{\sqrt{3}}$. The slope of the tangent is $m_t = -\frac{1}{\sqrt{3}}$. The slope of the normal is $m_n = -\frac{1}{m_t} = \sqrt{3}$. The equation of the normal at $P(1, \frac{2}{\sqrt{3}})$ is $y - \frac{2}{\sqrt{3}} = \sqrt{3}(x - 1)$. $\sqrt{3} y - 2 = 3x - 3 \Rightarrow 3x - \sqrt{3} y = 1$.

If $x+\sqrt{3} y=3$ is the tangent to the ellipse $2 x^2+3 y^2=k$ at a point $P$,then the equation of the normal to this ellipse at $P$ is

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