If the eccentricity of the standard hyperbola passing through the point is , then the equation of the tangent to the hyperbola at is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
For a standard hyperbola x²/a² - y²/b² = 1 with eccentricity e = 2, we have b² = a²(e²-1) = 3a².
Since (4,6) lies on the hyperbola: 16/a² - 36/3a² = 1, giving 16/a² - 12/a² = 4/a² = 1, so a² = 4 and b² = 12.
The tangent to x²/4 - y²/12 = 1 at point (4,6) is: (4x)/4 - (6y)/12 = 1, which simplifies to x - y/2 = 1, or 2x - y = 2, i.e., 2x - y - 2 = 0.
Since (4,6) lies on the hyperbola: 16/a² - 36/3a² = 1, giving 16/a² - 12/a² = 4/a² = 1, so a² = 4 and b² = 12.
The tangent to x²/4 - y²/12 = 1 at point (4,6) is: (4x)/4 - (6y)/12 = 1, which simplifies to x - y/2 = 1, or 2x - y = 2, i.e., 2x - y - 2 = 0.
[JEE Main 2019 · 8 Apr · Shift II]