Let A = {(x, y) : y = mx + 1},B = {(x, y) : x | vedclass

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(C) The line $y = mx + 1$ is tangent to the ellipse $x^2 + 4y^2 = 1$ if the condition for tangency $c^2 = a^2m^2 + b^2$ is satisfied.Here,the equation

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$-\frac{\sqrt{3}}{2}$

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(C) The line $y = mx + 1$ is tangent to the ellipse $x^2 + 4y^2 = 1$ if the condition for tangency $c^2 = a^2m^2 + b^2$ is satisfied.Here,the equation of the ellipse is $\frac{x^2}{1} + \frac{y^2}{1/4} = 1$,so $a^2 = 1$ and $b^2 = 1/4$.The line is $y = mx + 1$,so $c = 1$.Substituting these into the condition $c^2 = a^2m^2 + b^2$ gives $1^2 = (1)m^2 + \frac{1}{4}$.$1 = m^2 + \frac{1}{4} \implies m^2 = \frac{3}{4} \implies m = \pm \frac{\sqrt{3}}{2}$.For the point $(\alpha, \beta)$ to be in the first quadrant (where $\alpha > 0$),we analyze the intersection point. The point of tangency is given by $(x, y) = (-\frac{a^2m}{c}, \frac{b^2}{c})$.Substituting $a^2=1, b^2=1/4, c=1$,we get $x = -m$ and $y = 1/4$.Since we require $\alpha > 0$,we must have $-m > 0$,which implies $m Thus,$m = -\frac{\sqrt{3}}{2}$ is the only valid value.The sum of all possible values of $m$ is $-\frac{\sqrt{3}}{2}$.

Let $A = \{(x, y) : y = mx + 1\}$,$B = \{(x, y) : x^2 + 4y^2 = 1\}$,and $C = \{(\alpha, \beta) : (\alpha, \beta) \in A \text{ and } (\alpha, \beta) \in B \text{ and } \alpha > 0\}$. If set $C$ is a singleton set,then the sum of all possible values of $m$ is:

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