A line is a common tangent to the circle (x-3 | vedclass

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(A) Let the point of contact on the parabola $y^{2}=4x$ be $A(t^{2}, 2t)$.The equation of the tangent to the parabola at $A(t^{2}, 2t)$ is $ty = x + t

Vedclass · Accepted Answer

$9$

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(A) Let the point of contact on the parabola $y^{2}=4x$ be $A(t^{2}, 2t)$.The equation of the tangent to the parabola at $A(t^{2}, 2t)$ is $ty = x + t^{2}$,which can be written as $x - ty + t^{2} = 0$.Since this line is also tangent to the circle $(x-3)^{2} + y^{2} = 9$ with center $(3, 0)$ and radius $r=3$,the perpendicular distance from the center $(3, 0)$ to the line must be equal to the radius.$\left| \frac{3 - t(0) + t^{2}}{\sqrt{1^{2} + (-t)^{2}}} ight| = 3$$\left| 3 + t^{2} ight| = 3\sqrt{1 + t^{2}}$Squaring both sides: $(3 + t^{2})^{2} = 9(1 + t^{2})$$9 + t^{4} + 6t^{2} = 9 + 9t^{2}$$t^{4} - 3t^{2} = 0$$t^{2}(t^{2} - 3) = 0$Since the points lie in the first quadrant,$t > 0$,so $t^{2} = 3$,which gives $t = \sqrt{3}$.Thus,the point of contact on the parabola is $A(t^{2}, 2t) = (3, 2\sqrt{3})$. So,$a=3$ and $b=2\sqrt{3}$.The equation of the tangent line is $x - \sqrt{3}y + 3 = 0$.The point of contact on the circle $B(c, d)$ is the foot of the perpendicular from the center $(3, 0)$ to the tangent line $x - \sqrt{3}y + 3 = 0$.Using the formula for the foot of the perpendicular $(c, d)$ from $(x_{1}, y_{1})$ to $Ax + By + C = 0$:$\frac{c-3}{1} = \frac{d-0}{-\sqrt{3}} = -\frac{1(3) - \sqrt{3}(0) + 3}{1^{2} + (-\sqrt{3})^{2}} = -\frac{6}{4} = -\frac{3}{2}$$c - 3 = -\frac{3}{2} \Rightarrow c = 3 - \frac{3}{2} = \frac{3}{2}$$d = -\sqrt{3} 	imes (-\frac{3}{2}) = \frac{3\sqrt{3}}{2}$We need to find $2(a+c) = 2(3 + \frac{3}{2}) = 2(\frac{9}{2}) = 9$.

$A$ line is a common tangent to the circle $(x-3)^{2}+y^{2}=9$ and the parabola $y^{2}=4x$. If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant,then $2(a+c)$ is equal to ........ .

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