The point (0,3) is nearest to the c...
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The point (0,3) is nearest to the curve x2 = 2y at (a) (222,0) (b)(0,0) ©)(2,2) (d) none of these

JEE/Engineering Exams
Maths
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[ begin{array}{l} qquad begin{array}{c} x^{2}=2 y 2 x=2 frac{d y}{d x} Rightarrow frac{d y}{d x}=x end{array} & begin{array}{l} left(x_{1}, frac{x_{1}^{2}}{2}right) text { Let the pout al- curve }left(x^{2}=2 yright) text { be }left(left(x_{4}+8 xright)right) end{array} end{array} ] ( Rightarrow ) slope of temgent ( =x_{1} ) [ begin{array}{l} m_{1} m_{2}=-1 3left(frac{3-frac{x_{1}^{2}}{2}}{0-x_{1}}right)left(x_{1}right)=-1 end{array} ] ( Rightarrow quad 6-x_{1}^{2}=lambda quad Rightarrow quad x_{1}^{2}=84 Rightarrow x_{1}=2 ) Powt ( =left(2, frac{2^{2}}{2}right)=(2,2) ) Ophion c
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