<p>Find the equation of the tangent to the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ that is also tangent to the circle with diameter $AB$, where $A$ and $B$ are specific points on the hyperbola.</p>
Step-by-Step Solution
Key Concept: Use the general tangent form to a hyperbola and apply the tangency condition with the circle to find the slope $m$.
<p><strong>Step 1:</strong> The tangent to the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ has the form $y = mx ± \sqrt{9m^2 - 4}$.</p><p><strong>Step 2:</strong> For the tangent to also be tangent to the circle, we apply the tangency condition: $\frac{4m ± \sqrt{9m^2 - 4}}{\sqrt{1 + m^2}} = 4$.</p><p><strong>Step 3:</strong> Squaring and simplifying: $495m^4 + 104m^2 - 400 = 0$.</p><p><strong>Step 4:</strong> Solving the quadratic in $m^2$: $m^2 = \frac{4}{5}$, so $m = \frac{2}{\sqrt{5}}$.</p><p><strong>Step 5:</strong> Substituting back into the tangent equation: $5y = 2x + 4$.</p><p>∴ Answer is (b).</p>
Correct Answer: b