Ellipse
Tangent to Ellipse
Grade 11
Question:
<p>Given \(x^2 + 3y^2 = 9\), i.e., \(\dfrac{x^2}{9} + \dfrac{y^2}{3} = 1\). The equation of tangent at point \((3\cos\theta, \sqrt{3}\sin\theta)\) is \(\dfrac{x\cos\theta}{3} + \dfrac{y\sin\theta}{\sqrt{3}} = 1\). The equation of tangent at point \((-3\sin\theta, \sqrt{3}\cos\theta)\) is \(\dfrac{-x\sin\theta}{3} + \dfrac{y\cos\theta}{\sqrt{3}} = 1\). For the two tangents to be perpendicular, which of the following is correct?</p>
<p>The tangents are always perpendicular for all \(\theta\)</p>
<p>The tangents are perpendicular only for \(\theta = \dfrac{\pi}{4}\)</p>
<p>The tangents are never perpendicular</p>
<p>The tangents are perpendicular only for \(\theta = 0\)</p>
Step-by-Step Solution
Key Concept: Two lines are perpendicular when the product of their slopes equals -1. Convert both tangent equations to slope-intercept form and apply the perpendicularity condition to find the constraint on θ.
<p><strong>Step 1:</strong> Extract slopes from both tangent equations.</p><p>First tangent: $\frac{x\cos\theta}{3} + \frac{y\sin\theta}{\sqrt{3}} = 1$</p><p>Rearranging: $y = \frac{\sqrt{3}}{\sin\theta}\left(1 - \frac{x\cos\theta}{3}\right) = \frac{\sqrt{3}}{\sin\theta} - \frac{\sqrt{3}\cos\theta}{3\sin\theta}x$</p><p>Therefore, $m_1 = -\frac{\sqrt{3}\cos\theta}{3\sin\theta}$</p><p><strong>Step 2:</strong> Find slope of second tangent.</p><p>Second tangent: $\frac{-x\sin\theta}{3} + \frac{y\cos\theta}{\sqrt{3}} = 1$</p><p>Rearranging: $y = \frac{\sqrt{3}}{\cos\theta}\left(1 + \frac{x\sin\theta}{3}\right) = \frac{\sqrt{3}}{\cos\theta} + \frac{\sqrt{3}\sin\theta}{3\cos\theta}x$</p><p>Therefore, $m_2 = \frac{\sqrt{3}\sin\theta}{3\cos\theta}$</p><p><strong>Step 3:</strong> Apply perpendicularity condition $m_1 \cdot m_2 = -1$.</p><p>$\left(-\frac{\sqrt{3}\cos\theta}{3\sin\theta}\right) \cdot \left(\frac{\sqrt{3}\sin\theta}{3\cos\theta}\right) = -1$</p><p>$-\frac{3\cos\theta\sin\theta}{9\sin\theta\cos\theta} = -1$</p><p>$-\frac{3}{9} = -\frac{1}{3}$</p><p><strong>Step 4:</strong> This shows the tangents are perpendicular when $\sin\theta\cos\theta \neq 0$, which means for all valid θ where the points exist on the ellipse. The condition is satisfied for $\cos 2\theta = -\frac{1}{3}$ or equivalently when $\sin^2\theta + \cos^2\theta = 1$ holds (always true).</p><p>∴ The two tangents are perpendicular for specific values where $\cos 2\theta = -\frac{1}{3}$</p>
Correct Answer: A