Question

Locus of the point of intersection of two tangents to y² = 4ax which with the tangent at the vertex forms a triangle of constant area 4 sq. unit is

• A. x² (y² – 4ax) = 64
• B. x² (y² + 4ax) = 64
• C. x² (y² – 4ax) = 128
• D. x² (y² + 4ax) = 128

Answer 1

Answer: (A)

x² (y² – 4ax) = 64

Answer 2

Here, tangent yt = x + at² cuts y- axis at (0, at). Hence tangent at t₁ and t₂ make the intercept of

|a(t₂ –t₁)|.

Point of intersection of tangents are:

h = at₁t₂ , k = a (t₁+ t₂)

area of D = $\frac{1}{2}$|a² t₁t₂(t₁ – t₂) | = 4

̃ (at₁ t₂)² (a² (t₁ + t₂)² – 4a² t₁t₂) = 64

̃ h² (k² – 4ah) = 64

\ required locus ̃ x² (y² – 4ax) = 64