JEE Main 2022 · 24 June · morning · Single correct · +4 / −1
Let λ x - 2y = μ be a tangent to the hyperbola a²x² - y² = b². Then ( λ/a )² - ( μ/b )² is equal to :
- (a) -2
- (b) -4
- (c) 2
- (d) 4
Answer and solution
Answer (d) 4
Solution
Maths sir · 4 steps
Put the hyperbola in standard form, write the line as y = mx + c, and use the condition c² = a²m² − b².
- Standard form. Divide a²x² - y² = b² by b² to get x²/(b²/a²) - y²/b² = 1. The semi-axes squared are A² = b²/a² and B² = b².
- Slope form of line. Rewrite λ x - 2y = μ as y = λ/2x - μ/2. So m = λ/2 and c = -μ/2.
c² = A²m² - B² ⇒ μ²/4 = b²/a²·λ²/4 - b²
- Clear the fractions. Multiply by 4 to get μ² = b²λ²/a² - 4b². Divide by b²: μ²/b² = λ²/a² - 4.
- So. (λ/a)² - (μ/b)² = 4, which is option (d).
The idea to keep. For x²/A² - y²/B² = 1, the line y = mx + c is a tangent exactly when c² = A²m² - B². Always bring the curve to standard form first so you read A² and B² correctly.
Answer from the NTA final answer key. Solution by PrepWiser.