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AIEEE 2004

JEE Main 2004 Mathematics question on Conic Sections

If the line 2x+6y=22x + \sqrt{6}y = 2 touches the hyperbola x2−2y2=4x^2 - 2y^2 = 4, then the point of contact is
  1. A.
    (−2,6)(-2, \sqrt{6})
  2. B.
    (−5,26)(-5, 2\sqrt{6})
  3. C.
    (12,16)\left(\dfrac{1}{2}, \dfrac{1}{\sqrt{6}}\right)
  4. D.
    (4,−6)(4, -\sqrt{6})
Show answer and solution

Correct option: D

Solution

The condition for line y = mx + c to touch hyperbola is standard, but here use the point of contact formula directly.

Rewrite the line as: 2x + √6·y = 2, so the tangent to x²/4 - y²/2 = 1 at point (x₁, y₁) is:
x·x₁/4 - y·y₁/2 = 1

Comparing with 2x + √6·y = 2, i.e., x + (√6/2)y = 1:
x₁/4 = 1 → x₁ = 4
-y₁/2 = √6/2 → y₁ = -√6

So the point of contact is (4, -√6). Verify: 16 - 2(6) = 16 - 12 = 4 ✓, and 2(4) + √6(-√6) = 8 - 6 = 2 ✓.

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