- Difficulty
- Challenging
- Suggested time
- 7 min
- Marks
- 4
Differentiation rules practice question 19
Two distinct lines pass through the point and are tangent to the parabola . Find the slope of the steeper one.
Is on the parabola? If not, the slope at is no use: which unknown point should you name instead?
The point is not on the parabola. Let the tangency point be ; the tangent there has slope .
Write the tangent line at and demand that it pass through . That gives a quadratic in .
A whole number.
Theory UBC Math 100 §4.1CLP-1 §2.4Key results
Worked solution
Strategy
Parametrise the unknown point of tangency by its x-coordinate a. The tangent line at a is determined by the derivative; requiring it to pass through the given external point produces an equation whose solutions are the two tangency points.
Working
Slope y′(a) = 4a, through the point of tangency.
Substitute the external point (1, −6).
Two tangency points, as promised.
The line tangent at (3, 18) has slope 12.
Answer
The answer is 12.
Check
The line y = 12x − 18 passes through (1, −6), and 2x² − (12x − 18) = 2(x − 3)² has a double root, so it touches the parabola.
Where this goes wrong
Evaluating the derivative at x = 1 and answering 4. The line through (1, −6) with slope 4 is not tangent to the parabola; (1, −6) is not even on it.
Try a similar question
Find an equation of the tangent line to at the point where .Saved in this browser only.