JEE Main 202224 Jun, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let λx−2y=μ\lambda x - 2y = \muλx−2y=μ be a tangent to the hyperbola a2x2−y2=b2{a^2}{x^2} - {y^2} = {b^2}a2x2−y2=b2. Then (λa)2−(μb)2{\left( {{\lambda \over a}} \right)^2} - {\left( {{\mu \over b}} \right)^2}(aλ)2−(bμ)2 is equal to :AOption A. −-−2BOption B. −-−4COption C. 2DOption D. 4CheckSkip · show solution
