Q. 325.0( 2 Votes )

Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x,y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.

Answer :


A(0, a) and B(b, 0) are points on the Y-axis and X-axis respectively and P(x, y) is the midpoint of AB


Now the x-coordinate of the midpoint will be the addition of x coordinates of the points A and B divided by 2



b = 2x


Similarly, the y-coordinate



a = 2y


Hence coordinates of A and B are (0, 2y) and (2x, 0) respectively


Now AB is given as tangent to curve having P as point of contact


Slope of line given two points (x1, y1) and (x2, y2) on it is given by


Here (x1, y1) and (x2, y2) are A (0, 2y) and B(2x, 0) respectively


slope of tangent AB =


Hence slope of tangent is


Slope of tangent of a curve y = f(x) is given by




Integrate



log y = - log x + c


log y + log x = c


Using log a + log b = log ab


log xy = c …(a)


Now it is given that the curve is passing through (1, a)


Hence (1, 1) will satisfy the curve equation (a)


Putting values x = 1 and y = 1 in (a)


log1 = c


c = 0


Put c = 0 back in (a)


log xy = 0


xy = e0


xy = 1


Hence the equation of the curve is xy = 1


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