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Vector Algebra

Question
CBSEENMA12033077

Find the equation of a curve passing through the point (– 2, 3), given that the slope of the tangent to the curve at any point (x, y) is fraction numerator 2 straight y over denominator straight y squared end fraction.

Solution
We know that slope of tangent to a curve is given by dy over dx
From the given condition, 
                       dy over dx space equals space fraction numerator 2 straight x over denominator straight y squared end fraction
Separating the variables and integrating, 
                   integral straight y squared space dy space equals space 2 space integral space straight x space dx
therefore space space space space space space space space space space space space space straight y cubed over 3 space equals space straight x squared plus straight c                             ...(1)
Since curve passes through (-2, 3)
therefore space space space space space space space space space space space space space space space space fraction numerator left parenthesis 3 right parenthesis cubed over denominator 3 end fraction space equals space left parenthesis negative 2 right parenthesis squared plus straight c space space space space or space space space space space space 9 space equals space 4 plus straight c space space space rightwards double arrow space space space space straight c space equals space 5
Putting c = 5 in (1), we get,
            straight y cubed over 3 space equals space straight x squared plus 5 space space space or space space space straight y cubed space equals space 3 left parenthesis straight x squared plus 5 right parenthesis
which is required equation of curve.