A hyperbola passes through the point P(√2, √3) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point.

Let equation of hyperbola be x²/a² - y²/b² = 1
(ae, 0) = (±2, 0)
Point P lies on hyperbola
So, 2/a² - 3/b² = 1
Also b² = a²(e² - 1)
b² = 4 - a²
From (1) and (2)
2/a² - 3/(4 - a²) = 1
2(4 - a²) - 3a² = a²(4 - a²)
8 - 2a² - 3a² = 4a² - a⁴
a⁴ - 9a² + 8 = 0
(a² - 1)(a² - 8) = 0
a² = 1, a² = 8
Since a² ≠ 8
So, a² = 1, b² = 4 - 1 = 3
Hyperbola is x² - y²/3 = 1
dy/dx = 3x/y ⇒ dy/dx = √6 at P
Equation of tangent at P is y - √3 = √6(x - √2)
Option B (2√2, 3√3)
y - √3 = √6(x - √2)
3√3 - √3 = √6(2√2 - √2)
2√3 = √6
This is incorrect
Option C (3√2, 2√3)
y - √3 = √6(x - √2)
2√3 - √3 = √6(3√2 - √2)
√3 = 2√12 = 4√3
This is incorrect
Option A (2√2, 3√3)
3√3 - √3 = √6(2√2 - √2)
2√3 = √6(√2) = 2√3
Hence option A satisfies the equation.