Let H: -x²/a²+y²/b²=1 be the hyperbola, whose eccentricity is √3 and the length of the latus rectum is 4 √3. Suppose the point (α, 6), α>0 lies on H. If β is the product of the focal distances of the point (α, 6), then α²+β is equal to

Solution

Maths sir · 4 steps

The minus sits on x², so this hyperbola opens up and down and every formula runs on b.

1. Read the axis. In -x²/a²+y²/b²=1 the positive term is in y, so the transverse axis is the y-axis with semi-length b. The roles of a and b swap: a² = b²(e²-1), the latus rectum is 2a²/b, and the foci are (0, ± be).

   e² = 3 ⇒ a² = 2b², 2a²/b = 4b²/b = 4b = 4√3

2. Place the point. So b = √3, which gives b² = 3 and a² = 6. Putting (α, 6) on H gives -α²/6 + 36/3 = 1, so α²/6 = 11 and α² = 66.
3. Focal distances. The foci are (0, ± be) = (0, ± 3). The distance to (0, 3) is √(66 + 9) = √75 = 5√3, and the distance to (0, -3) is √(66 + 81) = √147 = 7√3. Their product is β = 35 × 3 = 105.
4. So. α² + β = 66 + 105 = 171, which is option (b).

The idea to keep. Before using any hyperbola formula, check which term is positive. That term's denominator sets the transverse axis, and e, the latus rectum and the foci are all measured along it.

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