Ellipse
Ellipse
nta_pyq_2025_jan
Grade 11
Question:
Let the product of the focal distances of the point (\sqrt3, 2 y on the ellipse , be . Then the 1 x 7 ) + = 1, (a > b) 2 a 2 b 2 4 absolute difference of the eccentricities of two such ellipses is 1-\sqrt3
\sqrt2 3-2\sqrt2
2\sqrt3 3-2\sqrt2
3\sqrt2 1-2\sqrt2
\sqrt3 2 2 2 2
Step-by-Step Solution
Key Concept: Apply the core result for ellipse parameters and tangents and simplify using the given constraints.
Product of focal distances = (a + ex1 ) (a - ex1 ) 2 2 2 2 2 (2) = a - e x 1 = a - e (3) 7 7 2 2 2 2 = a - 3e = \Rightarrow a = + 3e 4 4 2 2 \Rightarrow 4a = 7 + 12e 2 2 1 x y & (\sqrt3, ) lines on + = 1 2 a 2 b 2 3 1 \therefore + = 1 2 2 a 4b 3 1 + = 1 2 2 2 a 4 (a ) (1 - e ) 2 2 2 12 (1 - e ) + 1 = 4a (1 - e ) 2 2 2 13 - 12e = (7 + 12e ) (1 - e ) 2 2 2 4 \Rightarrow 13 - 12e = 7 - 7e + 12e - 12e 4 2 \Rightarrow 12e - 17e + 6 = 0 17 $\pm$ \sqrt289 - 288 17 $\pm$ 1 3 2 2 \therefore e = = = & 24 24 4 3 \sqrt3 2 \therefore e = &\sqrt 2 3 \sqrt3 2 3 - 2\sqrt 2 \therefore difference == - \sqrt = 2 3 2\sqrt 3
Correct Answer: 2