Question

If the normals from any point to the parabola y² = 4x cut the line x = 2 in points whose ordinates are in A.P., then the slopes of tangents at the co-normal points are in-

• A. H.P.
• B. G.P.
• C. A.P.
• D. None of these

Answer 1

Answer: (B)

G.P.

Answer 2

y² = 4x

eqⁿ of normal is y = – tx + 2t + t³ …(1)

it intersect x = 2 we get y = t³

let the three ordinates be $t_{1}^{3},t_{2}^{3},t_{3}^{3}$ in A.P.

2t² = $t_{1}^{3} + t_{3}^{3}$ ̃ (t₁ + t₃)³ – 3 t₁ t₃ (t₁ + t₃) …(2)

from eqⁿ (1) we have t₁ + t₂ + t₃ = 0

̃ t₁ + t₃ = – t₂

eqⁿ (2) reduces – $t_{2}^{3}$ – 3 t₁ t₃ (– t₂) = 2t³

̃ – $t_{2}^{3}$ + 3 t₁ t₂ t₃ = 2t³

3 $t_{2}^{3}$ = 3 t₁ t₂ t₃

$t_{2}^{2}$ = t₁ t₃ t₁, t₂, t₃ are in G.P.

So slope of tangent $\frac{1}{t_{1}},\frac{1}{t_{2}},\frac{1}{t_{3}}$ are in G.P.