Let E: x²/a²+y²/b²=1, a>b and H: x²/A²-y²/B²=1. Let the distance between the foci of E and the foci of H be 2 √3. If a-A=2, and the ratio of the eccentricities of E and H is 1/3, then the sum of the lengths of their latus rectums is equal to :

Solution

Maths sir · 4 steps

Same focal distance means ae = AE'. Use the eccentricity ratio there, and a and A drop out at once.

1. Read the condition. Both curves have their foci 2√3 apart, so each has half-focal distance √3. Let e and e' be the eccentricities of E and H. Then ae = √3 and Ae' = √3.

   ae = Ae' ⇒ a/A = e'/e = 3

2. Find a and A. Put a = 3A into a - A = 2 to get 2A = 2, so A = 1 and a = 3. Then e = √3/3 = 1/√3 and e' = √3. The ratio e : e' = 1 : 3 checks out.
3. Find b² and B². For the ellipse, b² = a² - (ae)² = 9 - 3 = 6. For the hyperbola, B² = (Ae')² - A² = 3 - 1 = 2.
4. So. The latus rectum of E is 2b²/a = 12/3 = 4, and the latus rectum of H is 2B²/A = 4/1 = 4. Their sum is 8, which is option (d).

The idea to keep. Focus distance is ae for an ellipse and Ae' for a hyperbola. Equal focal distances turn an eccentricity ratio into a ratio of semi-axes. Then use b² = a² - a²e² and B² = A²e'² - A².

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