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On the parabola y^2 = 4x, a normal chord is drawn at a point A so…
Question
On the parabola y^2 = 4x, a normal chord is drawn at a point A so that this chord subtends a right angle at the vertex O of the parabola. Find the square of the length of this normal chord.
✓ Verified answer: 108checked by our engine — not a guess
Step-by-step solution
Use the parametrisation A = (t^2, 2t) on y^2 = 4x (with a = 1). The normal at A meets the parabola again at the point with parameter t' = -t - 2/t.
The slope of OA is 2t/t^2 = 2/t, and of OB is 2/t'. For the chord AB to subtend a right angle at the vertex O:
(2/t)(2/t') = -1 => t t' = -4.
But t t' = t(-t - 2/t) = -t^2 - 2. So -t^2 - 2 = -4 => t^2 = 2, hence t = sqrt2.
Then t' = -sqrt2 - 2/sqrt2 = -sqrt2 - sqrt2 = -2 sqrt2.
A = (t^2, 2t) = (2, 2 sqrt2). B = (t'^2, 2t') = (8, -4 sqrt2).
(Check right angle: OA . OB = 2*8 + (2 sqrt2)(-4 sqrt2) = 16 - 16 = 0.)
Length^2 = (8-2)^2 + (-4 sqrt2 - 2 sqrt2)^2 = 6^2 + (-6 sqrt2)^2 = 36 + 72 = 108.
Final answer108
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