Previous year Question of jee mains
Chapter:- application and derivatives
Answers
EXPLANATION.
Tangent to the curve y = x³ at the point P(t, t³) meet the curve again at Q.
The ordinates of the point which divides PQ internally in the ratio 1 : 2.
As we know that,
⇒ P = (t, t³).
⇒ Q = (t₁, t₁³).
As we know that,
Slope of the line = (y₂ - y₁)/(x₂ - x₁).
Using this formula in the equation, we get.
Slope of the PQ = (t₁³ - t³)/(t₁ - t).
⇒ (t₁ - t)(t₁² + t₁t + t²)/(t₁ - t) = (t₁² + t₁t + t²).
Equation of the curve : y = x³.
dy/dx at point (t, t³) = 3t².
dy/dx = Slope of the tangent.
⇒ t₁² + t₁t + t² = 3t².
⇒ t₁² + t₁t + t² - 3t² = 0.
⇒ t₁² + t₍t - 2t² = 0.
⇒ (t₁ + 2t)(t₁ - t) = 0.
⇒ t₁ ≠ t.
⇒ t₁ = - 2t.
Co-ordinates of Q = (t₁, t₁³).
Co-ordinates of Q = (-2t, -8t³).
Ordinates of the point = (2t³ - 8t³)/2 + 1 = -6t³/3 = -2t³
Option [1] is correct answer.
- Tangent in the curve y = x³ at the point P(t,t³) meets the curve again at point Q.
- Ordinate of the point that divides PQ in the ratio of 1:2.
Slope of the tangent at point p
= (3x²)x = t = 3t²
So equation tangent at P(t,t³)
For point of intersection with y = x³
For x ≠ t
For Q : x = -2t, Q(-2t,-8t³)
Required Point
Ordinate of the point which the line PQ in the ratio of 1:2 ☞
- Option A.
Hope This Helps!!