The slopes of the focal chords of the parabol | vedclass

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(C) The equation of the circle is $x^2+y^2=2^2$,so the radius $r=2$.Any tangent to the circle with slope $m$ is given by $y=mx \pm r\sqrt{1+m^2}$,whic

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$\frac{1}{\sqrt{15}}, -\frac{1}{\sqrt{15}}$

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(C) The equation of the circle is $x^2+y^2=2^2$,so the radius $r=2$.Any tangent to the circle with slope $m$ is given by $y=mx \pm r\sqrt{1+m^2}$,which is $y=mx \pm 2\sqrt{1+m^2}$.The parabola is $y^2=32x$,which is of the form $y^2=4ax$ with $4a=32$,so $a=8$.The focus of the parabola is $(a, 0) = (8, 0)$.Since the tangent is a focal chord,it must pass through the focus $(8, 0)$.Substituting $(8, 0)$ into the tangent equation: $0 = m(8) \pm 2\sqrt{1+m^2}$.$8m = \mp 2\sqrt{1+m^2} \Rightarrow 4m = \mp \sqrt{1+m^2}$.Squaring both sides: $16m^2 = 1+m^2$.$15m^2 = 1 \Rightarrow m^2 = \frac{1}{15}$.Thus,$m = \pm \frac{1}{\sqrt{15}}$.

The slopes of the focal chords of the parabola $y^2=32x$,which are tangents to the circle $x^2+y^2=4$,are

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