JEE Main202131 Aug 2021Evening ShiftMathematicsEllipseActual
The locus of mid-points of the line segments joining - 3 , - 5 and the points on the ellipse x 2 4 + y 2 9 = 1 is :
Options
- A36 x 2 + 16 y 2 + 90 x + 56 y + 145 = 0
- B36 x 2 + 16 y 2 + 108 x + 80 y + 145 = 0
- C9 x 2 + 4 y 2 + 18 x + 8 y + 145 = 0
- D36 x 2 + 16 y 2 + 72 x + 32 y + 145 = 0
Correct answer
B. 36 x 2 + 16 y 2 + 108 x + 80 y + 145 = 0
Step-by-step solution
Parametric point on the given ellipse is 2 sin θ ,   3 cos θ Let, the mid-point of line segments joining - 3 , - 5 and 2 sin θ , 3 cos θ is h ,   k Then, by using mid-point formula, we get 2 sin θ - 3 2 = h ,   3 cos θ - 5 2 = k ⇒ 2 sin θ = 2 h + 3 ,   3 cos θ = 2 k + 5 ⇒ sin θ = 2 h + 3 2 ,   cos θ = 2 k + 5 3 We know, sin 2 θ + cos 2 θ = 1 ⇒ 2 h + 3 2 2 + 2 k + 5 3 2 = 1 ⇒ 1 4 4 h 2 + 9 + 12 h + 1 9 4 k 2