The equation of the common tangent touching t | vedclass

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(A) Let the equation of the tangent to the parabola $y^2 = 4x$ be $y = mx + \frac{a}{m}$,where $a = 1$. So,$y = mx + \frac{1}{m}$.This can be rewritte

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$y = x + 1$

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(A) Let the equation of the tangent to the parabola $y^2 = 4x$ be $y = mx + \frac{a}{m}$,where $a = 1$. So,$y = mx + \frac{1}{m}$.This can be rewritten as $mx - y + \frac{1}{m} = 0$.This line is also a tangent to the circle $(x-3)^2 + y^2 = 9$,which has center $(3, 0)$ and radius $r = 3$.The perpendicular distance from the center $(3, 0)$ to the line $mx - y + \frac{1}{m} = 0$ must be equal to the radius $3$.$\frac{|m(3) - 0 + \frac{1}{m}|}{\sqrt{m^2 + (-1)^2}} = 3$$|3m + \frac{1}{m}| = 3\sqrt{m^2 + 1}$Squaring both sides: $(3m + \frac{1}{m})^2 = 9(m^2 + 1)$$9m^2 + 6 + \frac{1}{m^2} = 9m^2 + 9$$\frac{1}{m^2} = 3 \implies m^2 = \frac{1}{3} \implies m = \pm \frac{1}{\sqrt{3}}$.Since the tangent is above the $X$-axis,we consider the positive slope $m = \frac{1}{\sqrt{3}}$.Substituting $m = \frac{1}{\sqrt{3}}$ into the tangent equation: $y = \frac{1}{\sqrt{3}}x + \frac{1}{1/\sqrt{3}} = \frac{x}{\sqrt{3}} + \sqrt{3}$.Multiplying by $\sqrt{3}$: $\sqrt{3}y = x + 3$,or $x - \sqrt{3}y + 3 = 0$.

The equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4x$ above the $X$-axis is

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