NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
A curve is such that the x -intercept of the tangent drawn to it at the point P x , y is reciprocal of the abscissa of P . Then, the equation of the curve is (where, c is the constant of integration and x > 1 )
Options
- Ay = c x 2 - 1
- By = c x 2 + 1
- Cy = c x 2 - 1
- Dy = c x - 1
Correct answer
C. y = c x 2 - 1
Step-by-step solution
The general equation of the tangent is Y - y = d y d x X - x ∴ x -intercept = x - y m ⇒ x - y m = 1 x ⇒ x - 1 x = y m Or m = y x - 1 x ⇒ d y d x = y x - 1 x ⇒ d y y = x d x x 2 - 1 On integrating, we get ∫ d y y = ∫ x x 2 - 1 d x ⇒ ln y = 1 2 ∫ 2 x d x x 2 - 1 = 1 2 ln x 2 - 1 + ln c ⇒ ln y = ln x 2 - 1 c ⇒ y = c ⋅ x 2 − 1