Question:
If the parabolas \(y^2 = 4b(x - c)\) and \(y^2 = 8ax\) have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)?
If the parabolas \(y^2 = 4b(x - c)\) and \(y^2 = 8ax\) have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)?
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Write the normal of each parabola in slope form, equate them and require \(m^2\gt 0\) (the trivial common normal \(m=0\), the x-axis, is excluded).
Updated On: Nov 5, 2026
- \((1, 1, 3)\)
- \((1, 1, 0)\)
- \((\frac{1}{2}, 2, 0)\)
- \((\frac{1}{2}, 2, 3)\)
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The Correct Option is A
Solution and Explanation
Normal to \(y^2=4Ax\) with slope \(m\): \(y = mx - 2Am - Am^3\).
For \(y^2=8ax\) (\(A=2a\)): \(y = mx - 4am - 2am^3\).
For \(y^2=4b(x-c)\) (\(A=b\), vertex shifted to \((c,0)\)): \(y = m(x-c) - 2bm - bm^3\).
For a common normal (\(m\ne0\)): \(-4a - 2am^2 = -c - 2b - bm^2\), so \(m^2(b-2a) = 4a - c - 2b\), i.e. \(m^2 = \frac{4a-c-2b}{b-2a}\).
We need \(m^2\gt0\).
(1) \((1,1,3)\): \(m^2 = \frac{4-3-2}{1-2} = 1\gt0\), valid.
(2) \((1,1,0)\): \(m^2 = \frac{4-0-2}{1-2} = -2\), invalid.
(3) \((\frac12,2,0)\): \(m^2 = \frac{2-0-4}{2-1} = -2\), invalid.
(4) \((\frac12,2,3)\): \(m^2 = \frac{2-3-4}{2-1} = -5\), invalid.
Answer: (1) \((1, 1, 3)\)
For \(y^2=8ax\) (\(A=2a\)): \(y = mx - 4am - 2am^3\).
For \(y^2=4b(x-c)\) (\(A=b\), vertex shifted to \((c,0)\)): \(y = m(x-c) - 2bm - bm^3\).
For a common normal (\(m\ne0\)): \(-4a - 2am^2 = -c - 2b - bm^2\), so \(m^2(b-2a) = 4a - c - 2b\), i.e. \(m^2 = \frac{4a-c-2b}{b-2a}\).
We need \(m^2\gt0\).
(1) \((1,1,3)\): \(m^2 = \frac{4-3-2}{1-2} = 1\gt0\), valid.
(2) \((1,1,0)\): \(m^2 = \frac{4-0-2}{1-2} = -2\), invalid.
(3) \((\frac12,2,0)\): \(m^2 = \frac{2-0-4}{2-1} = -2\), invalid.
(4) \((\frac12,2,3)\): \(m^2 = \frac{2-3-4}{2-1} = -5\), invalid.
Answer: (1) \((1, 1, 3)\)
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