Lesson 1.10 Tangents and Normals (Day 2)

Key Concepts

  • Tangent to a curve at point x = x0: line with slope m = \left.\dfrac{dy}{dx}\right|{x0} passing through (x0, y_0).

  • Normal: line perpendicular to tangent ⇒ slope mN = -\dfrac{1}{mT}.

  • Tangent from external point (xP, yP) to curve f(x) found by equating slope of line P(xP, yP)\text{–}T(x, f(x)) to derivative at T.

  • For a required slope m, solve \dfrac{dy}{dx} = m to find point(s) of contact.

Example Summary

• Example 1

  • Curve: y = 16 - x^2, line: y = -x + 4 intersects at A, B.

  • Find intersection; choose A; slope m = \left.\dfrac{dy}{dx}\right|A = -2xA.

  • Tangent: y - yA = m(x - xA).

• Example 2

  • Curve unspecified (implied): solve \dfrac{dy}{dx} = \text{slope of given line}.

  • Ensure tangent line shares that slope and point of contact.

• Example 3

  • External point: (4,0), curve y^2 = 4ax (general parabola form assumed).

  • Use equation of chord of contact or solve quadratic obtained from equal‐slope condition to find tangent points.

• Example 4

  • Tangent y = 2x touches y = x^3 + ax + b at x = 1.

  • Conditions:

    1. Contact point lies on curve: 1^3 + a(1) + b = 2(1).

    2. Slopes equal: \left.3x^2 + a\right|_{x=1} = 2.

  • Solve simultaneous equations for a, b.

• Example 5

  • External point (2,3) to y = x^2.

  • Let contact point (t,t^2).

  • Tangent slope at t: m = 2t.

  • Equation through external point: 3 - t^2 = (2t)(2 - t).

  • Solve quadratic in t to get both tangent points and equations.

Homework / Practice

  • Textbook p.388 Ex 16A: #3, 4, 7, 12, 14, 16.

  • Worksheet “Equations of Tangents and Normals” – circled questions only.

  • Handout: “More Tangents and Normals”.