Tangents are drawn from origin to the curve $y = \sin x + \cos x .$ Then their points of contact lie on the curve

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

$\frac{2}{x^{2}} - \frac{1}{y^{2}} = 1$

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

$\frac{2}{y^{2}} - \frac{1}{x^{2}} = 1$

answer is D.

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Detailed Solution

$y = \sin x + \cos x$

$\frac{d y}{d x} = \cos x - \sin x$

$y - y_{1} = (\cos x_{1} - \sin x_{1}) (x - x_{1}) ↑ (0, 0)$

$- y_{1} = (\cos x_{1} - \sin x_{1}) (- x_{1})$

$\cos x_{1} - \sin x_{1} = \frac{y_{1}}{x_{1}} \overset{}{\to} (1)$

$\cos x_{1} + \sin x_{1} = y \overset{}{\to} (2)$

${(1)}^{2} + {(2)}^{2} \Rightarrow$ $\frac{2}{y^{2}} - \frac{1}{x^{2}} = 1$

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