Question

PA and PB are the tangents drawn to y² = 4x from point P. These tangents meet the y-axis at the points A₁ and B₁ respectively. If the area of triangle PA₁ B₁ is 2 sq. units, then locus of ‘p’ is-

• A. a² (y² + 2x)x² = 8
• B. a² (y² + 4x)x² = 16
• C. a² (y² – 4x)x² = 8
• D. a² (y² – 4x)x² = 16

Answer 1

Answer: (D)

a² (y² – 4x)x² = 16

Answer 2

Let A ŗ (t₁², 2t₁), B ŗ (t₂², 2t₂), then P ŗ (t₁t₂, t₁ + t₂).

Also equations of PA and PB are yt₁ = x + at₁², yt₂ = x + at₂².

Thus, A₁ ŗ (0, at₁), B₁ ŗ (0, at₂) .

Now, area of triangle PA₁B₁ = $\frac{1}{2}$. |A₁B₁| |t₁ t₂|

= $\frac{1}{2}$a |t₁ – t₂| |t₁ t₂| = 2 (given)

Ž a² (t₁ – t₂)² (t₁t₂)² = 16

Ž a² [(t₁ + t₂)² – 4t₁t₂] (t₁t₂)² = 16

Thus locus of ‘P’ is a² (y² – 4x) x² = 16.