JEE Main 2025 · 24 January · morning · Single correct · +4 / −1
Let the product of the focal distances of the point (√3, 1/2) on the ellipse x²/a²+y²/b²=1,(a>b), be 7/4. Then the absolute difference of the eccentricities of two such ellipses is
- (a) (1-2 √2)/√3
- (b) (1-√3)/√2
- (c) (3-2 √2)/(2 √3)
- (d) (3-2 √2)/(3 √2)
Answer and solution
Answer (c) (3-2 √2)/(2 √3)
Solution
Maths sir · 4 steps
Focal distances are a ± ex, so their product gives one equation and the point itself gives the other.
- Product of focal distances. For a point (x₁, y₁) on the ellipse, the focal distances are a + ex₁ and a - ex₁. Their product is a² - e²x₁² = a² - 3e², and this must equal 7/4.
- Point on ellipse. Putting the point into the equation gives 3/a² + 1/4b² = 1, so b² = a²/4(a² - 3). Then e² = 1 - b²/a² = 1 - 1/4(a² - 3).
t = a² - 3: t + 3/4t = 7/4 ⇒ 4t² - 7t + 3 = 0
- Two ellipses. (t - 1)(4t - 3) = 0. With t = 1: a² = 4, b² = 1, e = √3/2. With t = 3/4: a² = 15/4, b² = 5/4, e = √(2/3). Both have a > b, so both are valid.
- So. √3/2 - √2/√3 = 3/2√3 - 2√2/2√3 = (3 - 2√2)/2√3, which is option (c).
The idea to keep. For a point on an ellipse, the focal distances are a + ex₁ and a - ex₁. Their sum is 2a and their product is a² - e²x₁², so you never need the distance formula.
Answer from the NTA final answer key. Solution by PrepWiser.