The parabolas y=x^(2)+1 and y=19-(x-a)^(2) are tangent to one another and tangent to the line y=mx+b. Find m and b.

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The parabolas  y=x^(2)+1 and  y=19-(x-a)^(2) are tangent to one another and tangent to the line  y=mx+b. Find  m and  b.

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Find Derivatives: We know that if two curves are tangent to each other at a point, they have the same slope at that point. The slope of a curve given by y=f(x) is the derivative f'(x). Let's find the derivatives of the given parabolas. First parabola: y = x^2 + 1 Derivative: dy/dx = 2x Second parabola: y = 19 - (x - a)^2 Derivative: dy/dx = -2(x - a)Set Equal Slopes: Since the parabolas are tangent to each other, their slopes are equal at the point of tangency. Therefore, we can set their derivatives equal to each other to find the x-coordinate of the point of tangency. 2x = -2(x - a) Solving for x gives us: 2x = -2x + 2a 4x = 2a x = a/2Calculate Tangency Point: Now we need to find the y-coordinate of the point of tangency by substituting x = a/2 into either of the original equations. We'll use the first parabola's equation. y = (a/2)^2 + 1 y = a^2/4 + 1Determine Line Equation: The line y=mx+b is tangent to both parabolas, which means it has the same slope as the parabolas at the point of tangency. Therefore, m must be equal to the derivative of the parabolas at the point of tangency. m = 2x = 2(a/2) = aFind Value of a: To find b, we need to use the point of tangency (a/2, a^2/4 + 1) and the slope m=a in the equation of the line y=mx+b. a^2/4 + 1 = a(a/2) + b Solving for b gives us: b = a^2/4 + 1 - a^2/4 b = 1Solve Quadratic Equation: We have found that m = a and b = 1. However, we still need to determine the value of a. To do this, we need to use the fact that the second parabola is also tangent to the line y=mx+b. We substitute y=mx+b into the second parabola's equation and solve for a. 19 - (x - a)^2 = ax + 1Calculate Discriminant: Expanding the equation and collecting like terms gives us: 19 - x^2 + 2ax - a^2 = ax + 1 -x^2 + (2a - a)x + (19 - a^2 - 1) = 0 -x^2 + ax + (18 - a^2) = 0Solve for a: Since the line is tangent to the parabola, this quadratic equation must have exactly one solution for x. This means the discriminant of the quadratic equation must be zero. Discriminant: (a)^2 - 4(-1)(18 - a^2) = 0 a^2 + 4(18 - a^2) = 0 a^2 + 72 - 4a^2 = 0 -3a^2 + 72 = 0Determine Possible Slopes: Solving for a gives us: -3a^2 = -72 a^2 = 24 a = ±√24 a = ±2√6 Since m = a, we have two possible values for m: m = 2√6 or m = -2√6.We have two possible slopes for the line, m = 2√6 or m = -2√6, and the y-intercept b = 1. These are the values of m and b for the line tangent to both parabolas.

The parabolas $y=x^2+1$ and $y=19-(x-a)^2$ are tangent to one another and tangent to the line $y=mx+b$ . Find $m$ and $b$ .

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The parabolas y=x2+1y=x^2+1y=x2+1 and y=19−(x−a)2y=19-(x-a)^2y=19−(x−a)2 are tangent to one another and tangent to the line y=mx+by=mx+by=mx+b. Find mmm and bbb.

The parabolas $y=x^2+1$ and $y=19-(x-a)^2$ are tangent to one another and tangent to the line $y=mx+b$ . Find $m$ and $b$ .