The line y=x+lambda is tangent to the ellipse | vedclass

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(C) The equation of the line is $y=x+\lambda$. On comparing it with $y=mx+c$, we get $m=1$ and $c=\lambda$. The equation of the ellipse is $2x^{2}+3y^

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$\sqrt{\frac{5}{6}}$

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(C) The equation of the line is $y=x+\lambda$. On comparing it with $y=mx+c$, we get $m=1$ and $c=\lambda$. The equation of the ellipse is $2x^{2}+3y^{2}=1$, which can be written as $\frac{x^{2}}{1/2} + \frac{y^{2}}{1/3} = 1$. On comparing it with $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, we get $a^{2}=\frac{1}{2}$ and $b^{2}=\frac{1}{3}$. If the line touches the ellipse, the condition for tangency is $c^{2}=a^{2}m^{2}+b^{2}$. Substituting the values, we get $\lambda^{2} = \frac{1}{2}(1)^{2} + \frac{1}{3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. Therefore, $\lambda = \pm \sqrt{\frac{5}{6}}$. Given the options, the correct value is $\lambda = \sqrt{\frac{5}{6}}$.

The line $y=x+\lambda$ is tangent to the ellipse $2x^{2}+3y^{2}=1$. Then, $\lambda$ is

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