Hyperbola
Normal to Hyperbola
Grade 11
Question:
<p>A normal to the hyperbola, \(4x^2 - 9y^2 = 36\) meets the co-ordinate axes \(x\) and \(y\) at \(A\) and \(B\), respectively. If the parallelogram \(OABP\) (\(O\) being the origin) is formed, then the locus of \(P\) is</p>
<p>\(4x^2 + 9y^2 = 121\)</p>
<p>\(9x^2 + 4y^2 = 169\)</p>
<p>\(4x^2 - 9y^2 = 121\)</p>
<p>\(9x^2 - 4y^2 = 169\)</p>
Step-by-Step Solution
Key Concept: A normal to the hyperbola at point (x₀, y₀) has a specific slope. The normal meets axes at A and B, forming a parallelogram OABP where P = A + B (vector addition), and eliminating the parameter gives the locus.
<p><strong>Step 1:</strong> Rewrite hyperbola in standard form: $\frac{x^2}{9} - \frac{y^2}{4} = 1$, so $a^2 = 9, b^2 = 4$.</p><p><strong>Step 2:</strong> For a point $(3\sec\theta, 2\tan\theta)$ on the hyperbola, the normal equation is: $\frac{3x}{\sec\theta} + \frac{2y}{\tan\theta} = 13$</p><p><strong>Step 3:</strong> Normal meets x-axis at $A(13\cos\theta/3, 0)$ (put $y=0$) and y-axis at $B(0, 13\sin\theta/2)$ (put $x=0$).</p><p><strong>Step 4:</strong> By parallelogram property, $P = A + B - O = (\frac{13\cos\theta}{3}, \frac{13\sin\theta}{2})$</p><p><strong>Step 5:</strong> Let $P = (h, k)$. Then $\cos\theta = \frac{3h}{13}$ and $\sin\theta = \frac{2k}{13}$</p><p><strong>Step 6:</strong> Using $\cos^2\theta + \sin^2\theta = 1$: $\frac{9h^2}{169} + \frac{4k^2}{169} = 1$</p><p><strong>Step 7:</strong> Therefore: $9x^2 + 4y^2 = 169$ (Ellipse)</p><p>∴ Answer: D</p>
Correct Answer: D