If P₁ and P₂ are two points on the ellipse x^2 4 +y^2=1 at which the tangents are parallel to the chord joining the points (0,1) and (2,0) , then the distance between P₁ and P₂ is

Step-by-step solution

Any tangent on an ellipse x^2 a^2 + y^2 b^2 =1 is given by y=m x a^2 m^2+b^2 Here a=2, b=1 aligned & m= 1-0 0-2 =- 1 2 & c= 4 (- 1 2 )^2+1^2 = 2 aligned So, y=- 1 2 x 2 For ellipse : x^2 4 + y^2 1 =1 We put y=- 1 2 x+ 2 aligned & x^2 4 + (- x 2 + 2 )^2=1 & x^2 4 + ( x^2 4 -2 ( x 2 ) 2 +2 )=1 & x^2+2 2 x+2=0 & or x^2-2 2 x+2=0 & x= 2 or - 2 aligned If x= 2 , y= 1 2 and x=- 2 , y=- 1 2 Points are ( 2 , 1 2 ), (- 2 ,- 1 2 ) . P₁ P₂= 1 2 - (- 1 2 ) ^2+ 2 -(- 2 ^2 . ) = ( 2 2 )^2+ (2 2 ^2 . = 2+8 = 10