AP EAMCET20228 Jul 2022Evening ShiftMathematicsEllipseActual
For belonging to an interval of length , suppose ( ,- ) is an interior point of the ellipse 4 x^2+5 y^2=1 . Then, (6 -4)²⁰¹+201=
Options
- A202
- B0
- C402
- D201
Correct answer
D. 201
Step-by-step solution
( ,- ) lies inside 4 x^2+5 y^2-1=0 aligned & 4 ^2+5 ^2-1 < 0 & 9 ^2-1 < 0 9 ^2 < 1 & ^2 < 1 9 - 1 3 < < 1 3 & (- 1 3 , 1 3 ) aligned Thus, = 1 3 + 1 3 = 2 3 Then, (6 -4)²⁰¹+201= (6 2 3 -4 )²⁰¹+201=201