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 is the differential equation?

(a) dy/dx = [(xy + 2) ± √(1 + xy)]/ x^2

(b) dy/dx = [(xy – 2) ± √(1 + xy)]/ x^2

(c) dy/dx = [(xy – 2) ± √(1 – xy)]/ x^2

(d) dy/dx = [(xy + 2) ± √(1 – xy)]/ x^2

This question was addressed to me in an online interview.

The query is from Linear First Order Differential Equations in portion Differential Equations of Mathematics – Class 12

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The correct choice is (c) dy/dx = [(xy – 2) ± √(1 – xy)]/ x^2

To explain: 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