JEE Main 2025April, Unspecified (session-level dataset)MathematicsMediumCoordinate Geometry478× since 2002The shortest distance between the curves y2=8xy^2 = 8xy2=8x and x2+y2+12y+35=0x^2 + y^2 + 12y + 35 = 0x2+y2+12y+35=0 is:AOption A. 23−12\sqrt{3} - 123−1BOption B. 2\sqrt{2}2COption C. 32−13\sqrt{2} - 132−1DOption D. 22−12\sqrt{2} - 122−1CheckSkip · show solution
