Geometric Applications
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).
Select the correct option:
Solution
y=2x
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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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- geometric applications
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
y=2x
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.
This easy difficulty mathematics question is from the chapter differential equations, covering the topic of geometric applications. It appeared in the 2025 exam.
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