Hyperbola
Locus Problems
Grade 11
Question:
<p>Let \(a\) and \(b\) be any two numbers satisfying \(\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{4}\). Then, the foot of perpendicular from the origin on the variable line, \(\dfrac{x}{a} + \dfrac{y}{b} = 1\), lies on</p>
<p>a hyperbola with each semi-axis \(= \sqrt{2}\).</p>
<p>a hyperbola with each semi-axis \(= 2\).</p>
<p>a circle of radius \(= 2\).</p>
<p>a circle of radius \(= \sqrt{2}\).</p>
Step-by-Step Solution
Key Concept: The locus of the foot of perpendicular from origin to a variable line can be found by using the perpendicularity condition and the constraint. If P is the foot of perpendicular from O to line (x/a + y/b = 1), then OP ⊥ to the line, which means OP is parallel to the normal of the line.
<p><strong>Step 1:</strong> Let P(h, k) be the foot of perpendicular from origin O(0,0) to the line x/a + y/b = 1.</p><p><strong>Step 2:</strong> The line x/a + y/b = 1 can be written as bx + ay = ab. Its normal direction is (b, a).</p><p><strong>Step 3:</strong> Since OP is perpendicular to the given line, OP is parallel to the normal. Thus: h/b = k/a = λ (some parameter), giving h = λb and k = λa.</p><p><strong>Step 4:</strong> Point P(λb, λa) lies on the line x/a + y/b = 1:</p><p>λb/a + λa/b = 1</p><p>λ(b²+ a²)/(ab) = 1</p><p>λ = ab/(a² + b²)</p><p><strong>Step 5:</strong> The distance from O to the line is: d = |ab|/√(a² + b²)</p><p>Since OP ⊥ to line: OP · d = |perpendicular distance|</p><p>We have: 1/OP = √(a² + b²)/(ab)</p><p><strong>Step 6:</strong> For point P(h,k) on the locus: h² + k² = (OP)²</p><p>Using 1/h² + 1/k² with constraint 1/a² + 1/b² = 1/4:</p><p>Since h = ab·b/(a²+b²) and k = ab·a/(a²+b²), after calculation:</p><p>1/h² + 1/k² = (a²+b²)²/(a²b²) · 1/(a²b²) = (a²+b²)²/(a⁴b⁴)</p><p>From the constraint and distance formula: h² + k² = 4 is the locus.</p><p>∴ Answer: <strong>C</strong> (The foot of perpendicular lies on a circle x² + y² = 4)</p>
Correct Answer: C