We are looking for straight lines that are tangent to both parabolas. These lines must have the form:
y = mx + b
The goal is to find the values of m and b so that the line is tangent to both parabolas
Step 1: Tangency Conditions
For a line y = mx + b to be tangent to a curve y = f(x), it must satisfy two conditions at the point of tangency x = a:
- The y-values must be equal: f(a) = ma + b
- The slopes must be equal: f'(a) = m
We'll apply this to both curves.
Step 2: Tangent to y = x² + 2
Let the line be tangent at x = a.
- The function value: a² + 2 = ma + b
- The derivative (slope): 2a = m
Substitute m = 2a into the first equation:
a² + 2 = 2a * a + b
a² + 2 = 2a² + b
Solving for b:
b = -a² + 2
Step 3: Tangent to y = -x² + 6x - 3
Let the line be tangent at x = c.
- The function value: -c² + 6c - 3 = mc + b
- The derivative (slope): -2c + 6 = m
Substitute m = -2c + 6 into the first equation:
-c² + 6c - 3 = (-2c + 6)c + b
-c² + 6c - 3 = -2c² + 6c + b
Subtract both sides:
c² - 3 - b = 0
Solving for b:
b = c² - 3
Step 4: Set Up a System
From the first curve:
- m = 2a
- b = -a² + 2
From the second curve:
- m = -2c + 6
- b = c² - 3
Since m and b must be the same for both:
- Set the slopes equal:
- 2a = -2c + 6 → a + c = 3
- Set the y-intercepts equal:
- -a² + 2 = c² - 3 → a² + c² = 5
Now substitute c = 3 - a into the second equation:
a² + (3 - a)² = 5
a² + 9 - 6a + a² = 5
2a² - 6a + 9 = 5
2a² - 6a + 4 = 0
Divide by 2: a² - 3a + 2 = 0
Solving the quadratic:
(a - 1)(a - 2) = 0 → a = 1 or a = 2
Step 5: Find the Equations of the Tangent Lines
If a = 1:
- m = 2a = 2
- b = -1² + 2 = 1
- Line: y = 2x + 1
If a = 2:
- m = 2a = 4
- b = -2² + 2 = -2
- Line: y = 4x - 2
Final Answer:
y = 2x + 1, y = 4x - 2
To verify these are the correct equations, we may graph the two parabolas and the two lines on Desmos. Notice that each line is tangent to both graphs simultaneously.
https://www.desmos.com/calculator/wskr7c3fa4
