Let Hₙ: x²/(1+n)-y²/(3+n)=1, n ∈ N. Let k be the smallest even value of n such that the eccentricity of Hₖ is a rational number. If l is the length of the latus rectum of Hₖ, then 21 l is equal to ____________.

Solution

Maths sir · 4 steps

Write e² as one fraction, then ask when both its top and bottom are perfect squares.

1. Eccentricity squared. Here a² = n+1 and b² = n+3. So e² = 1 + b²/a² = 1 + (n+3)/(n+1).

   e² = (2n+4)/(n+1) = 2(n+2)/(n+1)

2. When is e rational. For even n, n+1 is odd, so gcd(2n+4, n+1) = gcd(2, n+1) = 1 and the fraction is already in lowest terms. Then e is rational only if both n+1 and 2n+4 are perfect squares.

   Odd square n+1	n	2n+4	Perfect square?
   9	8	20	no
   25	24	52	no
   49	48	100	yes

3. Latus rectum. So k = 48, with a² = 49, a = 7, b² = 51 and e = 10/7. Then l = 2b²/a = (2 × 51)/7 = 102/7.
4. So. 21l = 21 × 102/7 = 3 × 102 = 306.

The common slip: 153. Taking the latus rectum as b²/a instead of 2b²/a drops the factor 2 and gives 153, not 306.

The idea to keep. For a square root of a fraction to be rational, reduce the fraction first. Once top and bottom share no common factor, each one must be a perfect square on its own.

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