The equation of a common tangent to the circl | vedclass

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(B) The equation of a tangent to the parabola $y^2 = 4ax$ (where $a = 1$) is $y = mx + \frac{a}{m}$,which is $y = mx + \frac{1}{m}$.This can be rewrit

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$\sqrt{3}y = x + 3$

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(B) The equation of a tangent to the parabola $y^2 = 4ax$ (where $a = 1$) is $y = mx + \frac{a}{m}$,which is $y = mx + \frac{1}{m}$.This can be rewritten as $m^2x - my + 1 = 0$.The circle is $x^2 + y^2 - 6x = 0$,which has center $(3, 0)$ and radius $r = 3$.Since the line is a tangent to the circle,the perpendicular distance from the center $(3, 0)$ to the line $m^2x - my + 1 = 0$ must equal the radius $3$.$\frac{|m^2(3) - m(0) + 1|}{\sqrt{(m^2)^2 + (-m)^2}} = 3$$|3m^2 + 1| = 3\sqrt{m^4 + m^2}$Squaring both sides: $(3m^2 + 1)^2 = 9(m^4 + m^2)$$9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$$3m^2 = 1$ $\Rightarrow m^2 = \frac{1}{3}$ $\Rightarrow m = \pm \frac{1}{\sqrt{3}}$.Taking $m = \frac{1}{\sqrt{3}}$,the equation is $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$,which simplifies to $\sqrt{3}y = x + 3$.

The equation of a common tangent to the circle $x^2 + y^2 - 6x = 0$ and the parabola $y^2 = 4x$ is:

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