Question

If the line 2x +$\sqrt{6}$y = 2 touches the hyperbola x² – 2y² = 4, then the point of contact is

• A. (–2, –$\sqrt{6}$)
• B. (–4, –$\sqrt{6}$)
• C. (4, –$\sqrt{6}$)
• D. (2, $\sqrt{6}$)

Answer 1

Answer: (C)

(4, –$\sqrt{6}$)

Answer 2

Equation of tangent to given hyperbola at (x₁, y₁) is xx₁ – 2yy₁= 4

On comparing this with 2x + $\sqrt{6}$y = 2

(x₁, y₁) ŗ (4, –$\sqrt{6}$)