Tangents are drawn to the hyperbola 4x² - y² = 36 at the points P and Q. If these tangents intersect at the point T(0, 3), then the area (in sq. units) of ΔPTQ is:

4x² - y² = 36
Equation of tangent at (x₁, y₁) is given by
xx₁/9 - yy₁/36 = 1
This passes through (0, 3) ⇒ 0 - 3y₁/36 = 1 ⇒ y₁ = -12
⇒ x₁²/9 - (-12)²/36 = 1
⇒ x₁²/9 - 1 = 1
⇒ x₁² = 18
⇒ x₁ = ±3√2
⇒ P(3√2, -12), Q(-3√2, -12), T(0, 3)
A = 1/2 | 3√2(-12 - 3) + (-3√2)(3 - (-12)) + 0(-12 - (-12)) |
= 1/2 | -45√2 - 45√2 |
= 45√2
Area = 1/2 * base * height
= 1/2 * 6√2 * 15 = 45√2