The equation of the common tangent to the cir | vedclass

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(A) Given parabola is $y^{2}=8x$,so $4a=8 \Rightarrow a=2$. Any tangent to the parabola is of the form $y=mx+\frac{a}{m}$,which is $y=mx+\frac{2}{m}$

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$y=x+2$

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(A) Given parabola is $y^{2}=8x$,so $4a=8 \Rightarrow a=2$. Any tangent to the parabola is of the form $y=mx+\frac{a}{m}$,which is $y=mx+\frac{2}{m}$ or $mx-y+\frac{2}{m}=0$. For this line to be a tangent to the circle $x^{2}+y^{2}=2$ (with center $(0,0)$ and radius $r=\sqrt{2}$),the perpendicular distance from the center to the line must equal the radius. $\frac{|m(0)-(0)+\frac{2}{m}|}{\sqrt{m^{2}+(-1)^{2}}}=\sqrt{2}$ $\frac{2}{|m|\sqrt{m^{2}+1}}=\sqrt{2}$ Squaring both sides: $\frac{4}{m^{2}(m^{2}+1)}=2 \Rightarrow m^{2}(m^{2}+1)=2$ $m^{4}+m^{2}-2=0$ $(m^{2}+2)(m^{2}-1)=0$ Since $m$ must be real,$m^{2}=1 \Rightarrow m=\pm 1$. For $m=1$,the tangent is $y=x+\frac{2}{1} \Rightarrow y=x+2$. For $m=-1$,the tangent is $y=-x+\frac{2}{-1} \Rightarrow y=-x-2$.

The equation of the common tangent to the circle $x^{2}+y^{2}=2$ and the parabola $y^{2}=8x$ is

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