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


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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Interactive Quiz on DIfferential CalculusFREE Class
Functional Equations - JEE with ease48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses