Question

A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?

(a) xy = 2

(b) xy = -1

(c) x – y = 2

(d) x + y = 2

This question was posed to me during a job interview.

My doubt stems from Linear First Order Differential Equations topic in portion Differential Equations of Mathematics – Class 12

Answer 1

The correct option is (d) x + y = 2

The best explanation: The equation of tangent to the curve y = f(x), at point (x, y), is

Y – y = dy/dx * (X – x) …..(1)

Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))

Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))

Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,

(x – y/(dy/dx))* (y – x/(dy/dx)) = 4

Or, (y – x/(dy/dx))^2 – 4dy/dx = 0

Or, x^2(dy/dx)^2 – 2(xy – 2)dy/dx + y^2 = 0

Solving for dy/dx we get,

dy/dx = [(xy – 2) ± √(1 – xy)]/ x^2

Let, 1 – xy = t^2

=> x(dy/dx) + y = -2t(dt/dx)

=> x^2(dy/dx) = t^2 – 1 – 2tx(dt/dx), so that (3) gives

t(x(dt/dx) – (t ± 1)) = 0

Hence, either t = 0

=>xy = 1 which is satisfied by (1, 1)

Or, x dt/dx = t ± 1

=> dx/x = dt/t ± 1

=> t ± 1 = cx

For x = 1, y = 1 and t = 0

=> c = ± 1, so the solution is

t = ± (x – 1) => t^2 = (x – 1)^2

Or, 1 – xy = x^2 – 2x + 1

Or, x + y = 2

Thus, the two curves that satisfies are xy = 1 and x + y = 2