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.

Question

Philoid · Accepted 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