JEE Main2016MathematicsApplication of DerivativesActual
If the tangent at a point P , with parameter t , on the curve x = 4 t 2 + 3 , y = 8 t 3 - 1 , t ∈ R , meets the curve again at a point Q , then the coordinates of Q are :
Options
- A16 t 2 + 3 , - 64 t 3 - 1
- B4 t 2 + 3 , - 8 t 3 - 1
- Ct 2 + 3 , t 3 - 1
- Dt 2 + 3 , - t 3 - 1
Correct answer
D. t 2 + 3 , - t 3 - 1
Step-by-step solution
Given P 4 t 2 + 3 ,   8 t 3 - 1 d y d t d x d t = d y d x = 3 t (slope of tangent at P ) Let Q = 4 λ 2 + 3 ,   8 λ 3 - 1 Slope of P Q = 3 t 8 t 3 - 8 λ 3 4 t 2 - 4 λ 2 = 3 t ⇒ t 2 + t λ - 2 λ 2 = 0 t - λ   t + 2 λ = 0 t = λ or λ = - t 2 ∴ Q t 2 + 3 ,   - t 3 - 1