Let the eccentricity of the hyperbola x²/a² - y²/b² = 1 be 5/4. If the equation of the normal at the point ( 8/√5,12/5 ) on the hyperbola is 8√5 x + β y = λ, then λ - β is equal to ___________.

Solution

Maths sir · 4 steps

Use e to link b² to a², put the point in to fix a², then write the normal and scale it.

1. Link b to a. For this hyperbola e² = 1 + b²/a². With e = 5/4, b²/a² = 25/16 - 1 = 9/16, so b² = 9/16a².
2. Use the point. Put x₁ = 8/√5, y₁ = 12/5 into the curve: 64/5a² - 144/25 × 16/9a² = (320 - 256)/25a² = 64/25a² = 1. So a² = 64/25 and b² = 9/16 × 64/25 = 36/25.

   (a² x)/x₁ + (b² y)/y₁ = a² + b²

3. Fill in the normal. a²/x₁ = 64/25 × √5/8 = 8√5/25, b²/y₁ = 36/25 × 5/12 = 15/25, and a² + b² = 100/25. Multiplying through by 25 gives 8√5 x + 15y = 100.
4. So. Matching with 8√5 x + β y = λ gives β = 15 and λ = 100, so λ - β = 85.

The common slip: 43. Using the ellipse normal a²x/x₁ - b²y/y₁ = a² - b² gives 8√5 x - 15y = 28, so λ - β = 28 + 15 = 43.

The idea to keep. Normal to x²/a² - y²/b² = 1 at (x₁, y₁) is a²x/x₁ + b²y/y₁ = a² + b². Both signs are plus for the hyperbola, unlike the ellipse.

Answer from the NTA final answer key. Solution by PrepWiser.