Ellipse
Perpendicular Radii
Grade 11
Question:
<p>C is the centre of the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) and A and B are two points on the ellipse such that \(\angle ACD = 90°\). Then \(\frac{1}{CA^2} + \frac{1}{CB^2} = \):</p>
<p>(a) \(\frac{25}{144}\)</p>
<p>(b) \(\frac{144}{25}\)</p>
<p>(c) \(\frac{7}{12}\)</p>
<p>(d) \(\frac{12}{7}\)</p>
Step-by-Step Solution
Key Concept: Use perpendicularity condition on vectors from centre to points on ellipse; apply algebraic identities to derive the sum.
<p>For ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\), we have \(a^2 = 16\), \(b^2 = 9\). Let A and B be points on the ellipse with \(\angle ACB = 90°\) where C is the centre. Parameterize: \(A = (4\cos\alpha, 3\sin\alpha)\), \(B = (4\cos\beta, 3\sin\beta)\). The perpendicularity condition \(\vec{CA} \cdot \vec{CB} = 0\) gives: \(16\cos\alpha\cos\beta + 9\sin\alpha\sin\beta = 0\). Then \(CA^2 = 16\cos^2\alpha + 9\sin^2\alpha\) and \(CB^2 = 16\cos^2\beta + 9\sin^2\beta\). Using the constraint and harmonic mean identity: \(\frac{1}{CA^2} + \frac{1}{CB^2} = \frac{1}{16} + \frac{1}{9} = \frac{9+16}{144} = \frac{25}{144}\).</p>
Correct Answer: A