Question

If the eccentricity of the hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] passing through the point \( (4, 6) \) is 2, then the equation of the tangent to this hyperbola at (4, 6) is

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

Hint:

When given a point on a hyperbola and the eccentricity, first use the identity \( e^2 = 1 + \frac{b^2}{a^2} \) to find parameters, and then apply the standard tangent formula.

Answer 1

The Correct Option is D

We are given: - Hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) - Eccentricity \( e = 2 \Rightarrow e^2 = 4 \) - From hyperbola identity: \( e^2 = 1 + \frac{b^2}{a^2} \Rightarrow 4 = 1 + \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = 3 \) Thus, \( \frac{b^2}{a^2} = 3 \Rightarrow \frac{y^2}{b^2} = \frac{y^2}{3a^2} \) So the equation becomes: \[ \frac{x^2}{a^2} - \frac{y^2}{3a^2} = 1 \Rightarrow \frac{1}{a^2}(x^2 - \frac{y^2}{3}) = 1 \Rightarrow x^2 - \frac{y^2}{3} = a^2 \] Substitute point (4, 6): \[ 4^2 - \frac{6^2}{3} = a^2 \Rightarrow 16 - \frac{36}{3} = a^2 \Rightarrow 16 - 12 = a^2 \Rightarrow a^2 = 4 \Rightarrow b^2 = 3a^2 = 12 \] So the hyperbola becomes: \[ \frac{x^2}{4} - \frac{y^2}{12} = 1 \] To find the tangent at (4,6), use the standard formula for tangent at point \( (x_0, y_0) \) on a hyperbola: \[ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \Rightarrow \frac{x . 4}{4} - \frac{y . 6}{12} = 1 \Rightarrow x - \frac{y}{2} = 1 \Rightarrow 2x - y = 2 \Rightarrow \boxed{2x - y - 2 = 0} \]