The equation of the common tangent to the cir | vedclass

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(B) For the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$. Any tangent to the parabola is given by $y = mx + \frac{a}{m}$,which simplifies to $mx -

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$y = x + 2$

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(B) For the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$. Any tangent to the parabola is given by $y = mx + \frac{a}{m}$,which simplifies to $mx - y + \frac{2}{m} = 0$. If this line is a tangent to the circle $x^2 + y^2 = 2$,the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $r = \sqrt{2}$. Thus,$\frac{|2/m|}{\sqrt{m^2 + 1}} = \sqrt{2}$. Squaring both sides,we get $\frac{4}{m^2} = 2(m^2 + 1)$,which leads to $2 = m^2(m^2 + 1)$. $m^4 + m^2 - 2 = 0$. Factoring gives $(m^2 + 2)(m^2 - 1) = 0$. Since $m^2$ must be positive,$m^2 = 1$,so $m = \pm 1$. Substituting $m = 1$ into the tangent equation,we get $y = x + \frac{2}{1} = x + 2$. Substituting $m = -1$,we get $y = -x - 2$.

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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