The equations of common tangents to the parab | vedclass

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(D) For the parabola $y^2=16x$,we have $a=4$. The equation of any tangent to the parabola is $y=mx+\frac{a}{m}$,which gives $y=mx+\frac{4}{m}$.For the

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$y=x+4, y=-x-4$

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(D) For the parabola $y^2=16x$,we have $a=4$. The equation of any tangent to the parabola is $y=mx+\frac{a}{m}$,which gives $y=mx+\frac{4}{m}$.For the circle $x^2+y^2=8$,the radius is $r=\sqrt{8}=2\sqrt{2}$. The equation of any tangent to the circle is $y=mx \pm r\sqrt{1+m^2}$,which gives $y=mx \pm 2\sqrt{2}\sqrt{1+m^2}$.Since these represent the same line,the constant terms must be equal:$\frac{4}{m} = \pm 2\sqrt{2}\sqrt{1+m^2}$.Squaring both sides,we get $\frac{16}{m^2} = 8(1+m^2)$.Dividing by $8$,we get $\frac{2}{m^2} = 1+m^2$,which implies $m^4+m^2-2=0$.Let $t=m^2$,then $t^2+t-2=0$,so $(t+2)(t-1)=0$. Since $t=m^2 \ge 0$,we have $m^2=1$,so $m=\pm 1$.Substituting $m=1$ into the tangent equation: $y=x+\frac{4}{1} \Rightarrow y=x+4$.Substituting $m=-1$ into the tangent equation: $y=-x+\frac{4}{-1} \Rightarrow y=-x-4$.Thus,the common tangents are $y=x+4$ and $y=-x-4$.

The equations of common tangents to the parabola $y^2=16x$ and the circle $x^2+y^2=8$ are

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