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A line is a common tangent to the circle\mathrm{ (x-3)^2+y^2=9} and the parabola \mathrm{y^2=4 x}. 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

10


Option: 2

9


Option: 3

8


Option: 4

7


Answers (1)

best_answer

\mathrm{\text { Let the coordinates of point } A \text { are }\left(t^2, 2 t\right) \text {. }}

Equation of tangent at point A is given by

\mathrm{y t=x+t^2 \Rightarrow x-t y+t^2=0}

Centre of circle is (3, 0).

\mathrm{\text { Also, radius }=\left|\frac{3-0+t^2}{\sqrt{1+t^2}}\right|=3}

\mathrm{\begin{aligned} & \Rightarrow \quad\left(3+t^2\right)^2=9\left(1+t^2\right) \\ & \Rightarrow 9+t^4+6 t^2=9+9 t^2 \\ & \Rightarrow t^4-3 t^2=0 \\ & \Rightarrow t^2\left(t^2-3\right)=0 \\ & \Rightarrow t=0,-\sqrt{3}, \sqrt{3} \end{aligned}}

\mathrm{\text { So, point } A(3,2 \sqrt{3}) \text { is in first quadrant, where } a=3 \text { and }b=2 \sqrt{3}}

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

\mathrm{\begin{aligned} & \therefore \quad \frac{c-3}{1}=\frac{d-0}{-\sqrt{3}}=\frac{-(3-0+3)}{4} \Rightarrow c=\frac{3}{2} \text { and } d=\frac{3 \sqrt{3}}{2} \\ & \therefore \quad 2(a+c)=2\left(3+\frac{3}{2}\right)=9 \end{aligned}}

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