The slope of the normal line to the curve x = t2 + 3t − 8 and y = 2t2 − 2t − 5…

Concept

When a curve is given in parametric form x = x(t), y = y(t), the slope of its tangent at a point is obtained from the chain rule as dy/dx = (dy/dt) / (dx/dt), valid wherever dx/dt is non-zero.

The normal at a point is the line through that point perpendicular to the tangent. Two perpendicular lines have slopes whose product is −1, so the slope of the normal is the negative reciprocal of the slope of the tangent: m(normal) = −1 / m(tangent).

Applying this to the given curve

1. Locate the parameter t of the given point. The value of t must satisfy both coordinate equations, so solve each one and keep only the value they share.

2. From the x-coordinate: t2 + 3t − 8 = 2, that is t2 + 3t − 10 = 0, which factorises as (t + 5)(t − 2) = 0 and gives t = −5 or t = 2.

3. From the y-coordinate: 2t2 − 2t − 5 = −1, that is 2t2 − 2t − 4 = 0, i.e. t2 − t − 2 = 0, which factorises as (t − 2)(t + 1) = 0 and gives t = 2 or t = −1.

4. The value shared by both lists is t = 2, so the point (2, −1) is the point of the curve at t = 2.

5. Differentiate each coordinate with respect to the parameter: dx/dt = 2t + 3 and dy/dt = 4t − 2.

6. Evaluate both derivatives at t = 2: dx/dt = 2(2) + 3 = 7 and dy/dt = 4(2) − 2 = 6.

7. Slope of the tangent: dy/dx = (dy/dt) / (dx/dt) = 6 / 7.

8. Slope of the normal: take the negative reciprocal of 6/7, giving −1 / (6/7) = −7/6.

Cross-check

• Substituting t = 2 back into the parametric equations returns x = 4 + 6 − 8 = 2 and y = 8 − 4 − 5 = −1, so t = 2 really does correspond to the stated point.

• The tangent slope and the normal slope multiply to (6/7) × (−7/6) = −1, which is the perpendicularity condition.

• Since dx/dt = 7 is non-zero at t = 2, the tangent is not vertical and dy/dx is defined there.

Slope of the normal line to the curve at (2, −1): −7/6.