Question:

<div class="solution"> <p><strong>Step 1:</strong> The equation of the ellipse is given by \(\frac{x^2}{65} + \frac{y^2}{14} = 1\). We can represent any point \(P\) on the ellipse using the parametric equations \(x = \sqrt{65} \cos\theta\) and \(y = \sqrt{14} \sin\theta\), where \(\theta\) is the parameter.</p> <p><strong>Step 2:</strong> The slope of the tangent at point \(P(\theta)\) can be found by differentiating the parametric equations with respect to \(\theta\). We have \(\frac{dx}{d\theta} = -\sqrt{65} \sin\theta\) and \(\frac{dy}{d\theta} = \sqrt{14} \cos\theta\). The slope of the tangent is given by \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = -\frac{\sqrt{14} \cos\theta}{\sqrt{65} \sin\theta}\). The slope of the normal at \(P(\theta)\) is the negative reciprocal of the slope of the tangent, which is \(\frac{\sqrt{65} \sin\theta}{\sqrt{14} \cos\theta}\).</p> <p><strong>Step 3:</strong> The equation of the normal at \(P(\theta)\) can be written as \(y - \sqrt{14} \sin\theta = \frac{\sqrt{65} \sin\theta}{\sqrt{14} \cos\theta} (x - \sqrt{65} \cos\theta)\). Simplifying, we get \(y - \sqrt{14} \sin\theta = \frac{\sqrt{65} \sin\theta}{\sqrt{14} \cos\theta} x - \frac{65 \sin\theta}{\sqrt{14} \cos\theta}\). Rearranging terms, the equation of the normal becomes \(y = \frac{\sqrt{65} \sin\theta}{\sqrt{14} \cos\theta} x - \frac{65 \sin\theta}{\sqrt{14} \cos\theta} + \sqrt{14} \sin\theta\).</p> <p><strong>Step 4:</strong> The normal intersects the ellipse again at \(Q(2\theta)\), which means the coordinates of \(Q\) satisfy both the equation of the ellipse and the equation of the normal. Substituting \(x = \sqrt{65} \cos(2\theta)\) and \(y = \sqrt{14} \sin(2\theta)\) into the equation of the normal, we get \(\sqrt{14} \sin(2\theta) = \frac{\sqrt{65} \sin\theta}{\sqrt{14} \cos\theta} \sqrt{65} \cos(2\theta) - \frac{65 \sin\theta}{\sqrt{14} \cos\theta} + \sqrt{14} \sin\theta\).</p> <p><strong>Step 5:</strong> Using the double angle identities \(\sin(2\theta) = 2\sin\theta \cos\theta\) and \(\cos(2\theta) = 2\cos^2\theta - 1\), and simplifying the equation obtained in Step 4, we can solve for \(\cos\theta\). After simplification and rearrangement, we arrive at an equation in terms of \(\cos\theta\) and \(\sin\theta\). Applying trigonometric identities and solving for \(\cos\theta\), we find \(\cos\theta = \frac{1}{3}\) as one of the solutions that satisfy the given conditions.</p> <p><strong>Answer:</strong> \(\cos\theta = \frac{1}{3}\)</p> <div class="key-concept"><strong>Key Concept:</strong> Parametric representation of an ellipse, equation of a normal to an ellipse, and trigonometric identities for solving the resulting equation.</div> </div>