question_answer

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 coordinate axes at A and B_ such that P is the mid-point of AB.

Answer:

Let the equation of the curve be y = f(x). The equation of the tangent to the curve at P(x, y) is \[Y-y=\frac{dy}{dx}(X-x).\] It is given that the tangent at P cuts the coordinate axes at A and B. The coordinate of A and B are \[\left( x-y\frac{dx}{dy},\,\,0 \right)\] and \[\left( 0,\,\,y\,-x\frac{dy}{dx},0 \right)\] respectively. It is given that P(x, y) is the mid-point of AB.  \[\therefore \] \[\frac{x-y\frac{dx}{dy}+0}{2}=x\] and \[\frac{0+y\frac{dy}{dx}+0}{2}=y\] \[\Rightarrow \] \[x-y\frac{dx}{dy}=2x\] and \[y-x\frac{dy}{dx}=2y\] \[\Rightarrow \] \[x=-\,y\frac{dx}{dy}\] and \[y=-\,x\frac{dy}{dx}\]             \[\Rightarrow \]   \[y=-\,x\frac{dy}{dx}\] \[\Rightarrow \]   \[\frac{1}{y}dy=-\frac{1}{x}dx\] \[\Rightarrow \]   \[\int{\frac{1}{y}dy}=-\int{\frac{1}{x}dx}\] \[\Rightarrow \]   \[\log \,y=-\log \,x+\log \,C\] \[\Rightarrow \]   \[\log \,xy=\log \,C\] \[\Rightarrow \]   xy = C                                      ?(i) This passes through (1, 1). Putting x = 1, y = 1 in Eq. (i). we get                         C = 1 Putting c = 1 in Eq. (i), we get                          xy = 1                 which is the required equation of curve.