Ellipse
Ellipse
nta_pyq_2025_jan
Grade 11
Question:
If the midpoint of a chord of the ellipse 2 y 2\sqrt\alpha is (\sqrt2, 4/3), and the length of the chord is , then \alpha is : x + = 1 9 4 3
Step-by-Step Solution
Key Concept: Apply the core result for ellipse parameters and tangents and simplify using the given constraints.
(2) 2 2 x y E : + = 1 9 4 T = S1 \sqrt2x 1 4 2 16 \Rightarrow + ( y) - 1 = + - 1 9 4 3 9 9(4) \sqrt2x y 2 4 + = + 9 3 9 9 \sqrt2x y 2 + = \Rightarrow \sqrt2x + 3y = 6 9 3 3 Now point of intersection of chord and ellipse is 2 2 (6 - 3y) y + = 1 18 4 2 2 (2 - y) y + = 1 2 4 2 2 2 (4 + y - 4y) + y = 4 2 \Rightarrow 3y - 8y + 4 = 0 2 \Rightarrow y = 2, 3 So, points are (0, 2) are (2\sqrt2, 2 3 ) 2 Length of chord = \sqrt(2\sqrt2) + ( 2 2 3 - 2) 16 = \sqrt8 + 9 \sqrt88 2\sqrt22 = = 3 3 On comparing \alpha = 22
Correct Answer: 2