Hyperbola
Normal to Hyperbola
Grade 11
Question:
<p>Given: \(4x^2 - 9y^2 = 36\). The equation of hyperbola is \(\frac{x^2}{9} - \frac{y^2}{4} = 1\). A point \(Q(3\sec\theta, 2\tan\theta)\) is on the hyperbola. The normal at \(Q\) meets the co-ordinate axes at \(A\left(\frac{13}{3}\sec\theta, 0\right)\) and \(B\left(0, \frac{13}{2}\tan\theta\right)\). If \(OABP\) is a parallelogram and coordinate of \(P\) is \((h, k)\), then the locus of \(P\) is:</p>
<p>\(9x^2 - 4y^2 = 196\)</p>
<p>\(9x^2 + 4y^2 = 169\)</p>
<p>\(9x^2 - 4y^2 = 196\)</p>
<p>\(9x^2 - 4y^2 = 169\)</p>
Step-by-Step Solution
Key Concept: Use the parametric form Q(3sec θ, 2tan θ) on the hyperbola, find the normal equation at Q, determine intercepts A and B on axes, then apply the parallelogram condition (diagonal bisection) to find P. The locus emerges by eliminating the parameter θ.
<p><strong>Step 1: Normal equation at Q(3sec θ, 2tan θ)</strong></p><p>For hyperbola <strong>x²/9 - y²/4 = 1</b>, the normal at point (x₀, y₀) is: <br><b>a²x/x₀ + b²y/y₀ = a² + b²</b><br>Here a² = 9, b² = 4, so: <b>9x/(3sec θ) + 4y/(2tan θ) = 13</b><br>Simplifying: <b>3x·cos θ + 2y·cot θ = 13</b></p><p><strong>Step 2: Find intercepts A and B</strong></p><p>At A (y = 0): <b>3x·cos θ = 13 ⟹ x = (13/3)sec θ</b> ✓ (matches given)<br>At B (x = 0): <b>2y·cot θ = 13 ⟹ y = (13/2)tan θ</b> ✓ (matches given)</p><p><strong>Step 3: Apply parallelogram condition OABP</strong></p><p>In parallelogram OABP with O at origin: <b>$\vec{OP} = \vec{OA} + \vec{OB}$</b><br>P = A + B = ((13/3)sec θ, 0) + (0, (13/2)tan θ)<br><b>P(h, k) = ((13/3)sec θ, (13/2)tan θ)</b><br>Therefore: h = (13/3)sec θ and k = (13/2)tan θ</p><p><strong>Step 4: Eliminate parameter θ</strong></p><p>From parametric equations:<br><b>sec θ = 3h/13</b> and <b>tan θ = 2k/13</b><br>Using identity <b>sec²θ - tan²θ = 1</b>:<br><b>(3h/13)² - (2k/13)² = 1</b><br><b>9h² - 4k² = 169</b></p><p>∴ <b>Answer: 9x² - 4y² = 169</b> (or equivalent form)</p>
Correct Answer: D