JEE Main202528 Jan 2025Evening ShiftMathematicsEllipseActual
If the midpoint of a chord of the ellipse x^2 9 + y^2 4 =1 is ( 2 , 4 / 3) , and the length of the chord is 2 3 , then is :
Options
- A20
- B22
- C18
- D26
Correct answer
B. 22
Step-by-step solution
aligned & E: x^2 9 + y^2 4 =1 & T=S₁ & 2 x 9 + 1 4 ( 4 3 y )-1= 2 9 + 16 9(4) -1 & 2 x 9 + y 3 = 2 9 + 4 9 & 2 x 9 + y 3 = 2 3 2 x +3 y=6 aligned Now point of intersection of chord and ellipse is aligned & (6-3 y)^2 18 + y^2 4 =1 & (2-y)^2 2 + y^2 4 =1 & 2 (4+y^2-4 y )+y^2=4 & 3 y^2-8 y+4=0 & y=2, 2 3 aligned So, points are (0,2) are (2 2 , 2 3 ) Length of chord = (2 2 )^2+ ( 2 3 -2 )^2 aligned & = 8+ 16 9 & = 88 3 = 2 22 3 aligned On comparing =22