Q.23 Find the equations of tangent and normal to the curve y = 2x³ - x² + 2 at point (1/2, 2) .

Answer: Tangent 2x - 4y + 7 = 0 ; normal 2x + y - 3 = 0 . dy/dx = 6x² - 2x . At x = 1/2 : slope of tangent = 6/4 - 1 = 1/2 . Tangent: y - 2 = 1/2(x - 1/2) ⇒ 4y - 8 = 2x - 1 ⇒ 2x - 4y + 7 = 0 . Slope of normal = -2 . Normal: y - 2 = -2(x - 1/2) ⇒ y - 2 = -2x + 1 ⇒ 2x + y - 3 = 0 .

Find the equations of tangent and normal to the curve \(y = 2x^3 - x^2 + 2\) at point \(\left(\dfrac12, 2\right)\).

Q.23 Find the equations of tangent and normal to the curve \(y = 2x^3 - x^2 + 2\) at point \(\left(\dfrac12, 2\right)\).

Doubtly Asked 28 Oct 2026 Board exam · March 2025 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: Tangent \(2x - 4y + 7 = 0\); normal \(2x + y - 3 = 0\).

\(\dfrac{dy}{dx} = 6x^2 - 2x\). At \(x = \frac12\): slope of tangent \(= \dfrac64 - 1 = \dfrac12\).

Tangent: \(y - 2 = \dfrac12\left(x - \dfrac12\right) \Rightarrow 4y - 8 = 2x - 1 \Rightarrow 2x - 4y + 7 = 0\).

Slope of normal \(= -2\). Normal: \(y - 2 = -2\left(x - \dfrac12\right) \Rightarrow y - 2 = -2x + 1 \Rightarrow 2x + y - 3 = 0\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?

Find the equations of tangent and normal to the curve \(y = 2x^3 - x^2 + 2\) at point \(\left(\dfrac12, 2\right)\).

1 answer

Your answer