The equation of a chord AB of an ellipse 2x^2 | vedclass

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(B) The equation of the ellipse is $2x^2 + y^2 = 1$. The equation of the chord is $x - y + 1 = 0$,which implies $y = x + 1$.Substituting $y = x + 1$ i

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$\operatorname{Tan}^{-1}(2)$

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(B) The equation of the ellipse is $2x^2 + y^2 = 1$. The equation of the chord is $x - y + 1 = 0$,which implies $y = x + 1$.Substituting $y = x + 1$ into the ellipse equation: $2x^2 + (x + 1)^2 = 1$.$2x^2 + x^2 + 2x + 1 = 1 \implies 3x^2 + 2x = 0$.$x(3x + 2) = 0$,so $x_1 = 0$ and $x_2 = -\frac{2}{3}$.Corresponding $y$ values: $y_1 = 0 + 1 = 1$ and $y_2 = -\frac{2}{3} + 1 = \frac{1}{3}$.Thus,the points are $A(0, 1)$ and $B(-\frac{2}{3}, \frac{1}{3})$.The slopes of $OA$ and $OB$ are $m_1 = \frac{1-0}{0-0} = \infty$ and $m_2 = \frac{1/3 - 0}{-2/3 - 0} = -\frac{1}{2}$.The angle $	heta$ between lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. Since $m_1 = \infty$,$	an 	heta = |\frac{1}{m_2}| = |\frac{1}{-1/2}| = 2$.Therefore,$	heta = \operatorname{Tan}^{-1}(2)$.

The equation of a chord $AB$ of an ellipse $2x^2 + y^2 = 1$ is $x - y + 1 = 0$. If $O$ is the origin,then $\angle AOB =$

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