Question

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is :

• A. \(2x - y - 2 = 0\)
• B. \(3x - 2y = 0\)
• C. \(2x - 3y + 10 = 0\)
• D. \(x - 2y + 8 = 0\)

Hint:

Use \(b^2 = a^2(e^2 - 1)\) to relate \(a\) and \(b\), fix \(a^2\) from the point, then write the tangent at \((4, 6)\).

Answer 1

The Correct Option is A

Let the hyperbola be \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). With \(e = 2\): \(b^2 = a^2(e^2 - 1) = 3a^2\).
Passing through \((4, 6)\): \(\frac{16}{a^2} - \frac{36}{3a^2} = 1\), so \(\frac{4}{a^2} = 1\), \(a^2 = 4\), \(b^2 = 12\).
Tangent at \((4, 6)\): \(\frac{4x}{4} - \frac{6y}{12} = 1\), i.e. \(x - \frac{y}{2} = 1\), so \(2x - y - 2 = 0\).
Answer: (1) \(2x - y - 2 = 0\)