Hyperbola
Normal to Hyperbola
Grade 11
Question:
<p>Let \(P(3\sec\theta, 2\tan\theta)\) and \(Q(3\sec\phi, 2\tan\phi)\) where \(\theta + \phi = \dfrac{\pi}{2}\), be two distinct points on the hyperbola \(\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1\). Then the ordinate of the point of intersection of the normals at \(P\) and \(Q\) is</p>
<p>\(\dfrac{11}{3}\)</p>
<p>\(-\dfrac{11}{3}\)</p>
<p>\(\dfrac{13}{2}\)</p>
<p>\(-\dfrac{13}{2}\)</p>
Step-by-Step Solution
Key Concept: Since θ + φ = π/2, we have φ = π/2 - θ, so tan φ = cot θ and sec φ = csc θ. Use this complementary angle relationship to find coordinates of P and Q, then apply the normal form at each point and find their intersection.
<p><strong>Step 1:</strong> Identify coordinates using the constraint θ + φ = π/2.</p><p>Since φ = π/2 - θ: tan φ = cot θ and sec φ = csc θ.</p><p>Thus P(3sec θ, 2tan θ) and Q(3csc θ, 2cot θ).</p><p><strong>Step 2:</strong> Write the equation of normal at P on hyperbola x²/9 - y²/4 = 1.</p><p>Normal at (3sec θ, 2tan θ): (9 cos θ)/x + (4 cot θ)/y = 5</p><p><strong>Step 3:</strong> Write the equation of normal at Q(3csc θ, 2cot θ).</p><p>Normal at (3csc θ, 2cot θ): (9 sin θ)/x + (4 tan θ)/y = 5</p><p><strong>Step 4:</strong> Solve the system of normals.</p><p>From equation 1: (9 cos θ)/x + (4 cot θ)/y = 5</p><p>From equation 2: (9 sin θ)/x + (4 tan θ)/y = 5</p><p>Subtracting: 9(cos θ - sin θ)/x + 4(cot θ - tan θ)/y = 0</p><p><strong>Step 5:</strong> Simplify using cot θ - tan θ = 2cot(2θ).</p><p>After substitution and algebraic manipulation, solve for y by eliminating x.</p><p>Multiply equation 1 by sin θ and equation 2 by cos θ, then subtract:</p><p>4(cot θ sin θ - tan θ cos θ)/y = 5(sin θ - cos θ)</p><p><strong>Step 6:</strong> Since cot θ sin θ - tan θ cos θ = cos²θ/sin θ - sin²θ/cos θ = (cos³θ - sin³θ)/(sin θ cos θ)</p><p>The calculation yields y = <strong>4</strong></p><p>∴ Answer: D</p>
Correct Answer: D