If the common tangents to the parabola x^2 = | vedclass

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(D) The equation of a tangent to the circle $x^2 + y^2 = 4$ with slope $m$ is $y = mx \pm 2\sqrt{1 + m^2}$.Since this line is also a tangent to the pa

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$3 + 2\sqrt{2}$

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(D) The equation of a tangent to the circle $x^2 + y^2 = 4$ with slope $m$ is $y = mx \pm 2\sqrt{1 + m^2}$.Since this line is also a tangent to the parabola $x^2 = 4y$,we substitute $y = mx + c$ into the parabola equation: $x^2 = 4(mx + c) \Rightarrow x^2 - 4mx - 4c = 0$.For tangency,the discriminant $D = 0$,so $(-4m)^2 - 4(1)(-4c) = 0$ $\Rightarrow 16m^2 + 16c = 0$ $\Rightarrow c = -m^2$.Comparing this with the circle's tangent form $c = \pm 2\sqrt{1 + m^2}$,we get $-m^2 = \pm 2\sqrt{1 + m^2}$.Squaring both sides: $m^4 = 4(1 + m^2) \Rightarrow m^4 - 4m^2 - 4 = 0$.Using the quadratic formula for $m^2$: $m^2 = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.Since $m^2$ must be positive,$m^2 = 2 + 2\sqrt{2}$.

If the common tangents to the parabola $x^2 = 4y$ and the circle $x^2 + y^2 = 4$ intersect at the point $P$,then find the square of the slope of the line.

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