# Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point.

Abscissa means the x-coordinate and ordinate means the y-coordinate

Slope of tangent is given as square of the difference of the abscissa and ordinate

Difference of abscissa and ordinate is (x – y) and its square will be (x – y)2

Hence slope of tangent is (x – y)2

Slope of tangent of a curve y = f(x) is given by  Substitute x – y = z hence y = x – z Differentiate x – z with respect to x    Using partial fraction for   Equating numerator

A(1 – z) + B(1 + z) = 1

Put z = 1

B = 1/2

Put z = -1

A = 1/2

Hence Put in (a) Integrate x = 1/2(log(1 + z) + (–log(1 – z)) + c

2x = log(1 + z) – log(1 – z) + c

Using log a – log b = log  Resubstitute z = x – y Now it is given that the curve is passing through origin that is (0, 0)

Hence (0, 0) will satisfy the curve equation (b)

Putting values x = 0 and y = 0 in (b) c = 0

Put c = 0 back in equation (b)   e2x(1 – x + y) = (1 + x – y)

Hence equation of curve is e2x(1 – x + y) = (1 + x – y)

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