Hyperbola
Tangent at a Point
Grade 11
Question:
<p>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</p>
<p>(a) 3<i>x</i> − 2<i>y</i> = 0</p>
<p>(b) <i>x</i> − 2<i>y</i> + 8 = 0</p>
<p>(c) 2<i>x</i> − <i>y</i> − 2 = 0</p>
<p>(d) 2<i>x</i> − 3<i>y</i> + 10 = 0</p>
Step-by-Step Solution
Key Concept: Use the eccentricity relation for hyperbola to find the ratio of a and b. Substitute the given point to find the actual values, then apply the tangent formula at a point.
<p><strong>Given:</strong> Standard hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ passes through (4, 6) with eccentricity <i>e</i> = 2</p><p><strong>For hyperbola:</strong> $e^2 = 1 + \frac{b^2}{a^2}$, so $4 = 1 + \frac{b^2}{a^2}$, giving $b^2 = 3a^2$</p><p><strong>Equation becomes:</strong> $\frac{x^2}{a^2} - \frac{y^2}{3a^2} = 1$</p><p><strong>Point (4, 6) on hyperbola:</strong> $\frac{16}{a^2} - \frac{36}{3a^2} = 1$</p><p>$\frac{16}{a^2} - \frac{12}{a^2} = 1 \Rightarrow a^2 = 4$, $b^2 = 12$</p><p><strong>Tangent at (4, 6):</strong> $\frac{4x}{4} - \frac{6y}{12} = 1$</p><p>$x - \frac{y}{2} = 1 \Rightarrow 2x - y - 2 = 0$</p><p>∴ Answer is <strong>(c) 2<i>x</i> − <i>y</i> − 2 = 0</strong></p>
Correct Answer: C