alleryEquation of normal to parabola y² = 4ax at (at², 2at):
Ay = mx + c
BBoth first and third
Cy - 2at = -t(x - at²)
Dy + tx = 2at + at³
Answer & Solution
Correct answer: B. Both first and third
At parameter t, slope of tangent = 1/t, so normal slope = -t.
**Normal:** y - 2at = -t(x - at²) → tx + y = 2at + at³.
Both A and C are correct equivalent forms.
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