Exams › JEE Main › Maths
Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is 2y/x². If the curve passes through the centre of the circle x² + y² - 2x - 2y = 0, then its equation is -
- x logₑ |y| = 2(x - 1)
- x logₑ |y| = 2(x - 1)
- x² logₑ |y| = -2(x - 1)
- x logₑ |y| = x - 1
Correct answer: x logₑ |y| = 2(x - 1)
Solution
The correct option is right because it satisfies the differential equation derived from the slope condition, and it also passes through the center of the given circle, which is (1, 1). This indicates that the relationship between x and y in the equation holds true for the specified conditions.
Related JEE Main Maths questions
- If √(1-x²ⁿ)+√(1-y²ⁿ)=a(xⁿ-yⁿ), then the value of (√(1-x²ⁿ) dy)/(√(1-y²ⁿ) dx) is
- For a curve that passes through the point (4, 0), the slope is governed by
dy/dx = y/x + 5x/((x + 2)(x − 3)).
If the point (5, a) lies on this curve, what is the value of a?
- Which differential equation represents the family of all conics whose axes are aligned with the coordinate axes?
- Find the equation of the curve that satisfies (xy - x²) (dy)/(dx) = y² and passes through the point (-1, 1).
- For the differential equation y = y/x + x/y, if its general solution is written as y = x / log|Cx|, then the function φ(x/y) is
- For the differential equation dy/dx = [y f'(x) − y²]/f(x), where f(x) is a specified function, the solution is
⚔️ Practice JEE Main Maths free + battle 1v1 →