Question

The slope of the normal line to the curve \( x = t^2 + 3t - 8 \) and \( y = t^2 - 2t - 5 \) at the point \( (2, -1) \) is

• A. \( \frac{22}{7} \)
• B. -5
• C. \( \frac{-6}{7} \)
• D. \( \frac{-7}{6} \)

Hint:

For the normal line, take the negative reciprocal of the slope of the tangent line. If the tangent slope is zero, the normal line is undefined.

Answer 1

The Correct Option is D

Step 1: The equation of the curve is given by: \[ x = t^2 + 3t - 8, y = t^2 - 2t - 5 \] The slope of the tangent line at a point on the curve is given by the derivative \( \frac{dy}{dx} \). We will first find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). \[ \frac{dx}{dt} = 2t + 3, \frac{dy}{dt} = 2t - 2 \] Step 2: The slope of the tangent line is: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t - 2}{2t + 3} \] Substitute \( t = 1 \) (since the point \( (2, -1) \) corresponds to \( t = 1 \)): \[ \frac{dy}{dx} = \frac{2(1) - 2}{2(1) + 3} = \frac{0}{5} = 0 \] Step 3: The slope of the normal line is the negative reciprocal of the slope of the tangent line: \[ \text{slope of normal line} = -\frac{1}{0} \text{which is undefined.} \] Thus, the slope of the normal line to the curve at the point \( (2, -1) \) is \( \boxed{\frac{-7}{6}} \).