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# The line y = –11x – 7 is tangent to the parabola y = ax² + bx + 1 at point P. The x-coordinate of P is –2. Determine the value of a + b.

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The line y = –11x – 7 is tangent to the parabola y = ax² + bx + 1 at point P. The x-coordinate of P is –2. Determine the value of a + b.

May 28, 2023

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The line y = –11x – 7 is tangent to the parabola y = ax² + bx + 1 at point P. This means that the line and the parabola share a common point, and the line's slope is equal to the parabola's slope at that point.

The slope of the line y = –11x – 7 is –11. The slope of the parabola y = ax² + bx + 1 at point P is equal to 2a + b.

Since the line and the parabola share a common point, and the line's slope is equal to the parabola's slope at that point, we have:

-11 = 2a + b

We are given that the x-coordinate of P is –2. This means that the point P is (–2, y).

We can substitute this point into the equation of the parabola to find the value of y:

y = a(–2)² + b(–2) + 1

y = 4a – 2b + 1

We are given that the line y = –11x – 7 is tangent to the parabola at point P. This means that the point P lies on the line y = –11x – 7.

We can substitute the point P into the equation of the line to find the value of y:

y = –11(–2) – 7

y = 15

We now have two equations with two unknowns:

-11 = 2a + b

15 = 4a – 2b

We can solve these equations for a and b.

Adding the two equations, we get:

4 = 6a

a = 2/3

Substituting a = 2/3 into the first equation, we get:

-11 = 4(2/3) + b

b = -19/3

Therefore, a + b = 2/3 + (-19/3) = -17/3.

May 28, 2023
#2
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Thank you so much for your thorough answer. Could you tell me how to solve this problem without using calculus?

somebodyIUsedToKnow  May 28, 2023
#3
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There's a mistake in the answer given above, the slope of the parabola should be 2ax + b, not 2a + b.

That leads to incorrect values for a and b.

Here's a solution not involving calculus.

To find out where the straight line and the parabola intersect equate the two expressions for y,

$$-11x-7=ax^{2}+bx+1,\\ \text{so} \\ ax^{2}+(b+11)x+8=0.$$

We need this to have equal roots at x = 2.

That gets you the two equations

$$-(b+11)/2a=2 \\ \text{and} \\ (b+11)^{2}-32a=0,$$

which can be solved simultaneously for a and b.

(a = 2, b = -19)

Tiggsy  May 28, 2023
#4
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Could you tell me why we need ax2 + (b+11)x + 8 = 0 to have equal roots at x = 2?

somebodyIUsedToKnow  May 28, 2023
edited by somebodyIUsedToKnow  May 28, 2023
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The equation will have either

1, no real roots, meaning that the line doesn't intersect with the parabola,

2, two real roots, meaning that the line intersects the parabola at two (different) points, or

3, two equal roots, meaning that the line contacts the parabola at a single point. In that case the line will be a tangent to the parabola.

Tiggsy  May 29, 2023