Let f(x)=x²+9, g(x)=x/(x-9) and a=f ° g(10), b=g ° f(3). If e and l denote the eccentricity and the length of the latus rectum of the ellipse x²/a+y²/b=1, then 8 e²+l² is equal to.

Solution

Maths sir · 4 steps

Get a and b from the compositions first, then remember they are the squares of the semi-axes.

1. Find a. g(10) = 10/(10-9) = 10, so a = f(10) = 100 + 9 = 109.
2. Find b. f(3) = 9 + 9 = 18, so b = g(18) = 18/(18-9) = 2. The ellipse is x²/109 + y²/2 = 1, with the major axis along the x-axis because 109 > 2.
3. Semi-axes. Here 109 and 2 are the squares of the semi-axes. The semi-major axis is √109 and the square of the semi-minor axis is 2.

   e² = 1 - 2/109 = 107/109, l = (2 × 2)/√109 = 4/√109

4. So. 8e² + l² = 856/109 + 16/109 = 872/109 = 8, which is option (c).

The idea to keep. For x²/A²+y²/B²=1 with A>B, use e² = 1 - B²/A² and latus rectum 2B²/A. The denominators are the squares of the semi-axes, so take square roots before using them.

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