The equation of the common tangent to the par | vedclass

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(D) The equation of a tangent to the parabola $y^2=8x$ (where $4a=8$,so $a=2$) is $y = mx + \frac{2}{m}$.This line is also a tangent to the circle $x^

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(D) The equation of a tangent to the parabola $y^2=8x$ (where $4a=8$,so $a=2$) is $y = mx + \frac{2}{m}$.This line is also a tangent to the circle $x^2+y^2=2$ (radius $r=\sqrt{2}$).The perpendicular distance from the center $(0,0)$ to the line $mx - y + \frac{2}{m} = 0$ must equal the radius $\sqrt{2}$.$\frac{|\frac{2}{m}|}{\sqrt{m^2+1}} = \sqrt{2}$ $\Rightarrow \frac{4}{m^2} = 2(m^2+1)$ $\Rightarrow m^4+m^2-2=0$.Let $t = m^2$,then $t^2+t-2=0 \Rightarrow (t+2)(t-1)=0$. Since $m^2 > 0$,we have $m^2=1$,so $m = \pm 1$.The tangent equation is $y = mx + \frac{2}{m}$. For $m=1$,$y = x + 2 \Rightarrow x - y + 2 = 0$. Here $a=1, b=-1$,so $-\frac{a}{b} = 1 > 0$.For $m=-1$,$y = -x - 2 \Rightarrow x + y + 2 = 0$. Here $a=1, b=1$,so $-\frac{a}{b} = -1 Thus,we take $a=1$ and $b=-1$.Then $3a^2+2b+1 = 3(1)^2 + 2(-1) + 1 = 3 - 2 + 1 = 2$.

The equation of the common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=2$ is $ax+by+2=0$. If $-\frac{a}{b} > 0$,then $3a^2+2b+1=$

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