Hyperbola
Focal Chords
Grade 11
Question:
<p>For a hyperbola \(9x^2 - 5y^2 = 1\), if P lies on the hyperbola at \((r_1\cos\theta, r_1\sin\theta)\) and Q lies on the hyperbola at \((r_2\cos(\theta+90°), r_2\sin(\theta+90°))\), find \(\frac{1}{r_1^2} + \frac{1}{r_2^2}\).</p>
Step-by-Step Solution
Key Concept: Substitute the polar-like coordinates into the hyperbola equation and use the fact that perpendicular radii vectors give complementary angles.
<p><strong>Step 1:</strong> P lies on \(9x^2 - 5y^2 = 1\):\[9r_1^2\cos^2\theta - 5r_1^2\sin^2\theta = 1\]</p><p><strong>Step 2:</strong> Dividing by \(r_1^2\):\[\frac{1}{r_1^2} = 9\cos^2\theta - 5\sin^2\theta\]</p><p><strong>Step 3:</strong> Q lies on \(9x^2 - 5y^2 = 1\):\[9r_2^2\sin^2\theta - 5r_2^2\cos^2\theta = 1\]</p><p><strong>Step 4:</strong> Dividing by \(r_2^2\):\[\frac{1}{r_2^2} = 9\sin^2\theta - 5\cos^2\theta\]</p><p><strong>Step 5:</strong> Adding:\[\frac{1}{r_1^2} + \frac{1}{r_2^2} = 9\cos^2\theta - 5\sin^2\theta + 9\sin^2\theta - 5\cos^2\theta = 4(\cos^2\theta + \sin^2\theta) = 4\]</p><p>∴ Answer is 4.</p>
Correct Answer: 4