Let the product of the focal distances of the point P(4,2 √3) on the hyperbola H: x²/a²-y²/b²=1 be 32 . Let the length of the conjugate axis of H be p and the length of its latus rectum be q. Then p²+q² is equal to__________

Solution

Maths sir · 4 steps

Turn the focal-distance product into e²x² - a², then pair it with the point lying on the curve.

1. Focal distances. For a point (x, y) on x²/a²-y²/b²=1, the distances to the foci are ex - a and ex + a. Their product is e²x² - a². With x = 4 and a²e² = a² + b², this gives 16e² - a² = 32.

   16(a²+b²)/a² - a² = 32 ⇒ b² = a²(a²+16)/16

2. Point on the curve. P(4, 2√3) lies on H, so 16/a² - 12/b² = 1. Put in b²: 16/a² - 192/a²(a²+16) = 1. Multiply by a²(a²+16) to get 16a² + 64 = a⁴ + 16a², so a⁴ = 64 and a² = 8.
3. Find b squared. b² = (8 × 24)/16 = 12. Check: 16/8 - 12/12 = 1, and e² = 20/8 = 2.5 gives 16 × 2.5 - 8 = 32.
4. So. p = 2b, so p² = 4b² = 48. q = 2b²/a = 24/2√2 = 6√2, so q² = 72. Then p² + q² = 48 + 72 = 120.

The common slip: 84. Taking the conjugate axis length as b instead of 2b gives p² = 12 and a total of 84. The full axis is 2b, so p² = 48.

The idea to keep. On a hyperbola the focal distances are ex ∓ a, so their product is e²x² - a². One condition from the foci plus one from the point lying on the curve fixes a² and b².

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