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  • A line is a common tangent to the circle  and the parabola

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 ______.
Option: 1 7
Option: 2 3
Option: 3 11
Option: 4 9

Answers (1)

best_answer

\text{Let the point of contact are }(t^2,2t)\;\;\;\text{)as a = 1)}

equation of the tangent at point A

\\\mathrm{yt}=\mathrm{x}+\mathrm{t}^{2} \\ \mathrm{x}-\mathrm{ty}+\mathrm{t}^{2}=0

center of the circle is (3,0) and radius is 3

the perpendicular distance from center to tangent is 3 unit, so

\\\left|\frac{3-0+\mathrm{t}^{2}}{\sqrt{1+\mathrm{t}^{2}}}\right|=3 \\ \left(3+\mathrm{t}^{2}\right)^{2}=9\left(1+\mathrm{t}^{2}\right) \\ 9+\mathrm{t}^{4}+6 \mathrm{t}^{2}=9+9 \mathrm{t}^{2} \\ \mathrm{t}=0,-\sqrt{3}, \sqrt{3} \\ \text { So point } \mathrm{A}(3,2 \sqrt{3})

since, point lies on first quadrant

\\\text{For point }\mathrm{B} \text{ which is foot of perpendicular from}\\ \text{centre (3,0) to the tangent }x-\sqrt{3} y+3=0\\ \frac{c-3}{1}=\frac{d-0}{-\sqrt{3}}=\frac{-(3-0+3)}{4}

\\\Rightarrow c=\frac{3}{2} \quad d=\frac{3 \sqrt{3}}{2} \\2(a+c) = 2\left(\frac{3}{2}+3\right)=9

 

 

Posted by

himanshu.meshram

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