Question

A line is tangent to one point and normal at other point on the curve x = 4t² + 3, y = 8t³ – 1 then slope of such line is

• A. ± 1
• B. – 1
• C. $\sqrt{2}$
• D. $\mp \sqrt{2}$

Answer 1

Answer: (D)

$\mp \sqrt{2}$

Answer 2

Let line is tangent at A(4t₁² + 3, 8t₁³ – 1) and normal at B(4t₂² + 3, 8t₂³ – 1)

⇒ $\left( \frac{dy}{dx} \right)_{A} = \frac{- 1}{(dy/dx)_{B}}$
= slope of line AB