AP EAMCET2011MathematicsStraight Lines
The locus of a point such that the sum of its distances from the points (0,2) and (0,-2) is 6 , is
Options
- A9 x^2-5 y^2=45
- B5 x^2+9 y^2=45
- C9 x^2+5 y^2=45
- D5 x^2-9 y^2=45
Correct answer
C. 9 x^2+5 y^2=45
Step-by-step solution
Let P (x₁, y₁ ) be any point, then aligned (x₁-0 )^2+ & (y₁-2 )^2 & + (x₁-0 )^2+ (y₁+2 )^2 =6 aligned aligned & x₁^2+ (y₁-2 )^2 =6- x₁^2+ (y₁+2 )^2 & x₁^2+ (y₁-2 )^2=36+ (x₁^2+ (y₁+2 )^2 ) & -12 x₁^2+ (y₁+2 )^2 & -8 y₁=36-12 x₁^2+ (y₁+2 )^2 & 3 x₁^2+ (y₁+2 )^2 =2 y₁+9 & 9 (x₁^2+ (y₁+2 )^2 )=4 y₁^2+81+36 y₁ & 9 x₁^2+5 y₁^2=45 aligned Hence, locus of a point is 9 x^2+5 y^2=45