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Which one of the following lines cannot be a normal to the parabola x² = 4y?
- x - y + 3 = 0
- x + y - 3 = 0
- x - 2y + 12 = 0
- x + 2y + 12 = 0
Correct answer: x - 2y + 12 = 0
Solution
The normal to x² = 4y has equation x + ty = 2t + t³ (at parameter t). Testing the lines, x - 2y + 12 = 0 cannot be expressed in normal form for any real t.
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