JEE PYQ: Parabola Question 22
Question 22 - 2019 (09 Jan Shift 1)
Equation of a common tangent to the circle, $x^2 + y^2 - 6x = 0$ and the parabola, $y^2 = 4x$, is:
(1) $2\sqrt{3}y = 12x + 1$
(2) $\sqrt{3}y = x + 3$
(3) $2\sqrt{3}y = -x - 12$
(4) $\sqrt{3}y = 3x + 1$
Show Answer
Answer: (2)
Solution
Tangent to $y^2 = 4x$: $y = mx + \frac{1}{m}$. Centre of circle $(3, 0)$, radius $= 3$. Distance condition: $\frac{|3m + \frac{1}{m}|}{\sqrt{1+m^2}} = 3$. Solving: $(3m^2+1)^2 = 9m^2(1+m^2) \Rightarrow m = \pm\frac{1}{\sqrt{3}}$. Tangent: $\sqrt{3}y = x + 3$.
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