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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

Equation of a common tangent to the circle, \(x^2 + y^2 - 6x = 0\) and the parabola, \(y^2 = 4x\) , is

  1. A \(2\sqrt 3 y = 12x + 1\)
  2. B \(\sqrt 3 y = x + 3\)
  3. C \(2\sqrt 3 y =  - x - 12\)
  4. D \(\sqrt 3 y = 3x + 1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt 3 y = x + 3\)

Step-by-step Solution

Detailed explanation

\(ty = x + {t^2}\) \(\left| {\frac{{3 + {t^2}}}{{\sqrt {1 + {t^2}} }}} \right| = 3\) \( \Rightarrow t = \sqrt 3 \) \( \Rightarrow \sqrt {3y} = x + 3\)
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