If a hyperbola passes through the point and has foci at , then the tangent to this hyperbola at also passes through the point
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
The foci are at (±2, 0), so c = 2, meaning a² + b² = 4 for a hyperbola x²/a² - y²/b² = 1 (with b² = 4 - a²).
Since P(√2, √3) lies on the hyperbola: 2/a² - 3/b² = 1. Substituting b² = 4 - a²:
2/a² - 3/(4 - a²) = 1
Let u = a²: 2(4 - u) - 3u = u(4 - u)
8 - 2u - 3u = 4u - u²
u² - 9u + 8 = 0
(u - 1)(u - 8) = 0
So a² = 1 (taking valid solution since a² < 4), giving b² = 3. The hyperbola is x² - y²/3 = 1.
The tangent at point (x₁, y₁) on x²/a² - y²/b² = 1 is: x·x₁/a² - y·y₁/b² = 1.
At P(√2, √3): x·√2/1 - y·√3/3 = 1, i.e., √2·x - (√3/3)·y = 1.
Check option 2, (2√2, 3√3): √2·(2√2) - (√3/3)·(3√3) = 4 - 3 = 1. ✓
Since P(√2, √3) lies on the hyperbola: 2/a² - 3/b² = 1. Substituting b² = 4 - a²:
2/a² - 3/(4 - a²) = 1
Let u = a²: 2(4 - u) - 3u = u(4 - u)
8 - 2u - 3u = 4u - u²
u² - 9u + 8 = 0
(u - 1)(u - 8) = 0
So a² = 1 (taking valid solution since a² < 4), giving b² = 3. The hyperbola is x² - y²/3 = 1.
The tangent at point (x₁, y₁) on x²/a² - y²/b² = 1 is: x·x₁/a² - y·y₁/b² = 1.
At P(√2, √3): x·√2/1 - y·√3/3 = 1, i.e., √2·x - (√3/3)·y = 1.
Check option 2, (2√2, 3√3): √2·(2√2) - (√3/3)·(3√3) = 4 - 3 = 1. ✓
[JEE Main 2017]