JEE Main 20202 Sep, Morning shiftMathematicsMediumCoordinate Geometry478× since 2002A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola x24−y22=1{{{x^2}} \over 4} - {{{y^2}} \over 2} = 14x2−2y2=1 at the point (x1,y1)\left( {{x_1},{y_1}} \right)(x1,y1). Then x12+5y12x_1^2 + 5y_1^2x12+5y12 is equal to :AOption A. 5BOption B. 6COption C. 10DOption D. 8CheckSkip · show solution
