IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Rohan Aggarwal

Q - a curve passes through (1,1) such that a triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has it's area equal to 2. Form the differential equation and find the equation of the curve ?

rahul.a

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

puneet

Now assuming that the curve is y = f(x), we get the equation of the tangent at any point (x,y) on curve as

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

So the point at which tangent meets X axis is ( (x(dy/dx) - y)/(dy/dx)) , 0)

and the point at which it meets Y axis is ( 0 , y - (dy/dx.)x )

So the area of triangle formed by the tangent and the co-ordinate axis is

= 1/2 * ((x(dy/dx) - y)/(dy/dx))*(y - (dy/dx.)x) = 2 (given)

So solving this equation we get a quadratic in dy/dx :

x² (dy/dx)² - (2xy - 4)dy/dx + y² = 0

Solving we get dy/dx = (2xy - 4 +

4(xy - 2)²- 4(xy)²) / 2 x²

& dy/dx = (2xy - 4 -

4(xy - 2)²- 4(xy)²) / 2 x²

Solve these equations by putting xy = t and then using the boundary condition i.e the curve passes through (1,1).

cheers ......... let us know if u stuck somewhere.

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