|
|
Okay, so I need to figure out how to solve this problem where I have to find the value of 'a' such that the line y = 3x + 2 is tangent to the parabola y = ax² + 4x + 5. Hmm, tangency condition... I remember that for a line to be tangent to a parabola, they must intersect at exactly one point. That means when I set the equations equal to each other, the resulting quadratic equation should have exactly one solution. Which implies the discriminant has to be zero. Let me try to work this out step by step.
First, let me set the two equations equal since the points of intersection satisfy both equations. So:
3x + 2 = ax² + 4x + 5
Now, let's rearrange this equation to bring all terms to one side. Subtract 3x and 2 from both sides:
0 = ax² + 4x + 5 - 3x - 2
Simplify the like terms. 4x - 3x is x, and 5 - 2 is 3. So:
0 = ax² + x + 3
So the quadratic equation is ax² + x + 3 = 0. For this quadratic to have exactly one real solution, the discriminant must be zero. The discriminant D of a quadratic ax² + bx + c is D = b² - 4ac. Let's compute that.
Here, a is the coefficient of x², which is 'a' (from the parabola equation), b is the coefficient of x, which is 1, and c is the constant term, which is 3. So:
D = (1)² - 4 * a * 3 = 1 - 12a
For the quadratic to have exactly one real solution, set D = 0:
1 - 12a = 0
Now, solve for 'a':
1 = 12a
Divide both sides by 12:
a = 1/12
Wait, let me check that again. If I have 1 - 12a = 0, then 12a = 1, so a = 1/12. Yeah, that seems right. Let me verify by plugging this back into the quadratic equation and see if it results in a perfect square.
If a = 1/12, then the quadratic becomes (1/12)x² + x + 3 = 0. Multiply both sides by 12 to eliminate the fraction:
x² + 12x + 36 = 0
This factors into (x + 6)² = 0, which indeed has a double root at x = -6. So the line touches the parabola at x = -6. Let me check if this point actually lies on both the line and the parabola.
First, substitute x = -6 into the line equation y = 3x + 2:
y = 3*(-6) + 2 = -18 + 2 = -16
Now, substitute x = -6 into the parabola equation with a = 1/12:
y = (1/12)(-6)^2 + 4(-6) + 5 = (1/12)*36 + (-24) + 5 = 3 - 24 + 5 = -16
Perfect, both give y = -16. So the point of tangency is (-6, -16), and the value of 'a' is indeed 1/12. Therefore, the answer should be a = 1/12.
But wait, let me just make sure I didn't mix up any coefficients. Let me go through the steps again.
Original equations:
Line: y = 3x + 2
Parabola: y = ax² + 4x + 5
Set equal: 3x + 2 = ax² + 4x + 5
Subtract 3x + 2: 0 = ax² + x + 3
Discriminant: b² - 4ac = (1)^2 - 4a3 = 1 - 12a
Set discriminant to zero: 1 - 12a = 0 => a = 1/12
Yes, that's consistent. And verification shows it works. So I think that's correct. Maybe I can also think about the derivative approach. If the line is tangent to the parabola, then at the point of tangency, their slopes are equal. The slope of the line is 3. The derivative of the parabola is y' = 2ax + 4. Setting that equal to 3:
2ax + 4 = 3
Solve for x:
2ax = -1 => x = -1/(2a)
So the x-coordinate of the point of tangency is -1/(2a). Now, this x should also satisfy both equations (the line and the parabola). Let's substitute x = -1/(2a) into the line equation:
y = 3*(-1/(2a)) + 2 = -3/(2a) + 2
And substitute into the parabola equation:
y = a*(-1/(2a))² + 4*(-1/(2a)) + 5 = a*(1/(4a²)) - 4/(2a) + 5 = (1/(4a)) - 2/a + 5
Set these two expressions for y equal:
-3/(2a) + 2 = (1/(4a)) - 2/a + 5
Let's solve this equation for 'a'. Multiply all terms by 4a to eliminate denominators:
4a*(-3/(2a)) + 4a2 = 4a(1/(4a)) - 4a*(2/a) + 4a*5
Simplify each term:
First term: 4a*(-3/(2a)) = -6
Second term: 4a*2 = 8a
Third term: 4a*(1/(4a)) = 1
Fourth term: -4a*(2/a) = -8
Fifth term: 4a*5 = 20
So the equation becomes:
-6 + 8a = 1 - 8 + 20
Simplify the right side:
1 - 8 is -7, then -7 + 20 is 13
So:
-6 + 8a = 13
Add 6 to both sides:
8a = 19
Divide by 8:
a = 19/8
Wait, that's conflicting with the previous result. Now I'm confused. Which one is correct?
Hold on, this is a problem. Using the discriminant method, I got a = 1/12, but using the derivative method, I ended up with a = 19/8. That can't be right. There must be a mistake here. Let me check the derivative approach again.
So, the derivative of the parabola y = ax² + 4x + 5 is y' = 2ax + 4. The slope of the tangent line is 3, so set 2ax + 4 = 3. Solving for x gives x = (3 - 4)/(2a) = (-1)/(2a). So x = -1/(2a). That part is correct.
Now, substitute this x into the line equation y = 3x + 2:
y = 3*(-1/(2a)) + 2 = -3/(2a) + 2
Substitute into the parabola equation:
y = a*( (-1/(2a)) )² + 4*(-1/(2a)) + 5
Calculate each term:
First term: a*(1/(4a²)) = 1/(4a)
Second term: 4*(-1/(2a)) = -2/a
Third term: 5
So total y = 1/(4a) - 2/a + 5
Set equal to the line's y:
-3/(2a) + 2 = 1/(4a) - 2/a + 5
Multiply all terms by 4a:
Left side: 4a*(-3/(2a)) + 4a*2 = (-6) + 8a
Right side: 4a*(1/(4a)) + 4a*(-2/a) + 4a*5 = 1 - 8 + 20a
Wait, here's the mistake! When multiplying the right side, the term 4a5 is 20a, not 20. Because 4a5 = 20a. But in my previous calculation, I wrote 20 instead of 20a. That's where the error is.
So correcting that:
Right side after multiplying by 4a:
1 - 8 + 20a
So the equation is:
-6 + 8a = 1 - 8 + 20a
Simplify the right side:
1 - 8 = -7, so -7 + 20a
Thus:
-6 + 8a = -7 + 20a
Now, bring variables to one side and constants to the other:
Subtract 8a from both sides:
-6 = -7 + 12a
Add 7 to both sides:
1 = 12a
Therefore, a = 1/12
Ah, there we go! So the derivative method also gives a = 1/12, which matches the discriminant method. I must have made a mistake in my initial derivative approach by forgetting to multiply the last term by 'a'. Now that it's corrected, both methods agree. So that confirms that a = 1/12 is the correct answer.
Just to recap: Both methods lead to a = 1/12. The discriminant method by ensuring the quadratic equation has exactly one solution, and the derivative method by ensuring the slopes are equal at the point of tangency. Both confirm the same result. Additionally, substituting back into the original equations shows that the point (-6, -16) lies on both the line and the parabola when a = 1/12, and the slope at that point is indeed 3, matching the line's slope.
Therefore, I'm confident that the correct value of 'a' is 1/12.
To determine the value of ( a ) such that the line ( y = 3x + 2 ) is tangent to the parabola ( y = ax^2 + 4x + 5 ), we need to ensure they intersect at exactly one point. This requires the quadratic equation formed by setting them equal to have a discriminant of zero.
Set the equations equal:
[
3x + 2 = ax^2 + 4x + 5
]
Rearrange terms:
[
ax^2 + x + 3 = 0
]
Calculate the discriminant:
For ( ax^2 + x + 3 = 0 ), the discriminant ( D ) is:
[
D = b^2 - 4ac = (1)^2 - 4(a)(3) = 1 - 12a
]
Set ( D = 0 ) for tangency:
[
1 - 12a = 0 \implies a = \frac{1}{12}
]
Verification using derivatives:
The slope of the line is 3.
The derivative of the parabola ( y = ax^2 + 4x + 5 ) is ( y' = 2ax + 4 ).
Set the derivative equal to 3 to find the point of tangency:
[
2ax + 4 = 3 \implies x = -\frac{1}{2a}
]
Substitute ( x = -\frac{1}{2a} ) into both equations and solve for ( a ), confirming ( a = \frac{1}{12} ).
Check the point of tangency:
Substitute ( a = \frac{1}{12} ) into the quadratic equation:
[
\frac{1}{12}x^2 + x + 3 = 0 \implies x^2 + 12x + 36 = 0 \implies (x + 6)^2 = 0
]
The point ( x = -6 ), ( y = -16 ) lies on both the line and the parabola, confirming tangency.
Final Answer: (\boxed{\dfrac{1}{12}})
|
|
发表于 2025-11-15 13:39:11