- Difficulty
- Capstone
- Suggested time
- 10 min
- Marks
- 5
Differentiation rules practice question 22
There is exactly one line that is tangent to both of the parabolas below, at different points. Find its slope.
The line touches each parabola at its own point, and the two points differ. How many unknowns does that call for, and what must two descriptions of one line agree on?
Let the line touch the first parabola at and the second at . Write both tangent lines in slope-intercept form.
One line, so the two slopes agree and the two intercepts agree. That is two equations in and .
A number, as a fraction or a decimal.
Theory UBC Math 100 §4.1CLP-1 §2.6Key results
Worked solution
Strategy
A common tangent is the same line written two ways. Write the tangent to each parabola at an unknown point, then equate slopes and equate intercepts.
Working
Derivative 2x; tangent at (a, a²).
Derivative 2x − 4; tangent at (b, b² − 4b + 5).
Equal slopes.
Equal intercepts, factored as a difference of squares.
Using b − a = 2. The line is y = x/2 − 1/16, touching at x = 1/4 and x = 9/4.
Answer
The answer is one half.
Check
x² − (x/2 − 1/16) = (x − 1/4)² and x² − 4x + 5 − (x/2 − 1/16) = (x − 9/4)², so the line touches each parabola once.
Where this goes wrong
Assuming the line touches both parabolas at the same x, which forces a = b and no solution. The two points of tangency are different, which is why two unknowns are needed.
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