Geometric Applications

Question

At every point on a curve the slope equals the ratio of the y-coordinate to the x-coordinate, and we want the equation of the curve passing through the point (1, 2).

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The geometric condition that the slope at each point equals the ratio of the coordinates translates directly into the differential equation \frac{dy}{dx} = \frac{y}{x}, a separable equation. The insight is that a slope equal to y/x means the tangent line always points along the radial direction from the origin, which characterizes straight lines through the origin. Separating gives \frac{dy}{y} = \frac{dx}{x}, and integrating both sides yields \ln|y| = \ln|x| + C_1. Exponentiating produces y = Cx, the family of lines through the origin. Applying the point (1, 2) gives 2 = C\cdot 1, so C = 2 and the curve is y = 2x. Option y = x^2 + 1 has slope 2x, which does not equal y/x. Option y = x + 1 fails the radial-slope condition since its slope is constant. Option y = 2/x has negative slope and does not pass through (1,2) with the required tangent direction. This is the standard JEE Advanced geometric-application setup. As a final check, for y = 2x the slope is 2 and y/x = 2x/x = 2 at every point, so the condition holds identically and the point (1,2) lies on the line.

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