IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

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Lets say curve has the equation y = f(x)

Now the equation of tangent at any point on the curve is given by

Y - y = (dy/dx)(X - x)

Now say, this line meets x axis at point (a,0) ...so the points where it meets y axis is (0,4/a) ( As the area of triangle is 4 )

By putting these two points in the equation above and eliminating the value of 'a' we get --

x²dy - xydy -ydx + 4dx = 0

x⁴( x²dy - 2xydy)/x⁴ + xydy - ydx + 4dx = 0

x⁴ d(y/x²) + (xy - y + 4)dx = 0

x⁴ . y/x² + y. x²/2 - xy + 4x = C

This curve passes through (1,1)..... By putting the this value in curve we get the final equation of the curve

3 x²y - 2 x y + 8 x - 9 = 0 Answer

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