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