If the length of the latus rectum of the ellipse x²+4 y²+2 x+8 y-λ=0 is 4 , and l is the length of its major axis, then λ+l is equal to ____________.

Solution

Maths sir · 4 steps

Complete the squares first, and then both the latus rectum and the major axis depend on λ + 5.

1. Complete the squares. x² + 2x = (x+1)² - 1 and 4y² + 8y = 4(y+1)² - 4. So the equation becomes (x+1)² + 4(y+1)² = λ + 5.
2. Standard form. Divide by λ + 5 to get (x+1)²/(λ+5) + (y+1)²/((λ+5)/4) = 1. So a² = λ + 5 and b² = (λ+5)/4, with a > b, so the major axis is horizontal.

   2b²/a = (2 · (λ+5)/4)/√(λ+5) = √(λ+5)/2 = 4

3. Find lambda and l. √(λ+5) = 8, so λ + 5 = 64 and λ = 59. Then a = 8, and the major axis is l = 2a = 16.
4. So. λ + l = 59 + 16 = 75.

The common slip: 67. Taking the major axis as a = 8 instead of 2a = 16 gives 59 + 8 = 67, not 75.

The idea to keep. For a shifted ellipse, complete the squares and read a² and b² from the standard form. The latus rectum is 2b²/a and the major axis is 2a, not a.

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