Question:
Minimum distance between the curves \(y^2 = 4x\) and \(x^2 + y^2 - 12x + 31 = 0\) is
Minimum distance between the curves \(y^2 = 4x\) and \(x^2 + y^2 - 12x + 31 = 0\) is
Show Hint
Minimum distance occurs along common normal.
Updated On: Apr 23, 2026
- \(\sqrt{5}\)
- \(\sqrt{21}\)
- \(\sqrt{28} - \sqrt{5}\)
- \(\sqrt{21} - \sqrt{5}\)
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \text{Circle: } (x-6)^2 + y^2 = 5, \text{ center } (6,0), r = \sqrt{5} \]
Step 2: Calculation / Simplification}
Parabola: \(y^2 = 4x\)
Normal to parabola at \((t^2, 2t)\): \(y = -tx + 2t + t^3\)
Normal passes through \((6,0)\): \(0 = -6t + 2t + t^3 \Rightarrow t^3 - 4t = 0\)
\(t(t^2 - 4) = 0 \Rightarrow t = 0, \pm 2\)
Points on parabola: \((0,0)\) and \((4, \pm 4)\)
Distance from \((6,0)\) to \((0,0)\) = 6
Distance from \((6,0)\) to \((4, \pm 4)\) = \(\sqrt{(6-4)^2 + (0 \mp 4)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}\)
Minimum distance = \(2\sqrt{5} - \sqrt{5} = \sqrt{5}\)
Step 3: Final Answer
\[ \sqrt{5} \]
\[ \text{Circle: } (x-6)^2 + y^2 = 5, \text{ center } (6,0), r = \sqrt{5} \]
Step 2: Calculation / Simplification}
Parabola: \(y^2 = 4x\)
Normal to parabola at \((t^2, 2t)\): \(y = -tx + 2t + t^3\)
Normal passes through \((6,0)\): \(0 = -6t + 2t + t^3 \Rightarrow t^3 - 4t = 0\)
\(t(t^2 - 4) = 0 \Rightarrow t = 0, \pm 2\)
Points on parabola: \((0,0)\) and \((4, \pm 4)\)
Distance from \((6,0)\) to \((0,0)\) = 6
Distance from \((6,0)\) to \((4, \pm 4)\) = \(\sqrt{(6-4)^2 + (0 \mp 4)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}\)
Minimum distance = \(2\sqrt{5} - \sqrt{5} = \sqrt{5}\)
Step 3: Final Answer
\[ \sqrt{5} \]
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