Math Problem Statement

Use implicit differentiation to find the points where the parabola defined by x2 - 2xy + y2 + 2x - 6y + 21 = 0 has horizontal and vertical tangent lines. The parabola has horizontal tangent lines at the point(s) The parabola has vertical tangent lines at the point(s)

Solution

We are tasked with finding the points on the given parabola where there are horizontal and vertical tangent lines using implicit differentiation.

Given equation of the parabola:

x22xy+y2+2x6y+21=0x^2 - 2xy + y^2 + 2x - 6y + 21 = 0

Step 1: Implicit Differentiation

We differentiate both sides of the equation with respect to xx. Keep in mind that yy is a function of xx, so apply the chain rule where necessary.

  1. Differentiating x2x^2:
    ddx(x2)=2x\frac{d}{dx}(x^2) = 2x

  2. Differentiating 2xy-2xy:
    Using the product rule:
    ddx(2xy)=2(y+xdydx)\frac{d}{dx}(-2xy) = -2 \left( y + x\frac{dy}{dx} \right)

  3. Differentiating y2y^2:
    ddx(y2)=2ydydx\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}

  4. Differentiating 2x2x:
    ddx(2x)=2\frac{d}{dx}(2x) = 2

  5. Differentiating 6y-6y:
    ddx(6y)=6dydx\frac{d}{dx}(-6y) = -6\frac{dy}{dx}

  6. Differentiating the constant 21:
    ddx(21)=0\frac{d}{dx}(21) = 0

Step 2: Combine all differentiated terms

2x2(y+xdydx)+2ydydx+26dydx=02x - 2 \left( y + x\frac{dy}{dx} \right) + 2y \frac{dy}{dx} + 2 - 6 \frac{dy}{dx} = 0

Simplify the equation:

2x2y2xdydx+2ydydx+26dydx=02x - 2y - 2x \frac{dy}{dx} + 2y \frac{dy}{dx} + 2 - 6 \frac{dy}{dx} = 0

Group the terms with dydx\frac{dy}{dx} and the rest:

(2x2y+2)+(2x+2y6)dydx=0(2x - 2y + 2) + \left( -2x + 2y - 6 \right) \frac{dy}{dx} = 0

Step 3: Solve for dydx\frac{dy}{dx}

(2x+2y6)dydx=(2x2y+2)\left( -2x + 2y - 6 \right) \frac{dy}{dx} = -(2x - 2y + 2)

dydx=(2x2y+2)2x+2y6\frac{dy}{dx} = \frac{-(2x - 2y + 2)}{-2x + 2y - 6}

Simplifying the expression:

dydx=2x2y22x2y+6\frac{dy}{dx} = \frac{2x - 2y - 2}{2x - 2y + 6}

Step 4: Conditions for Horizontal and Vertical Tangents

  • Horizontal Tangent Line occurs when dydx=0\frac{dy}{dx} = 0. Set the numerator equal to zero:

    2x2y2=02x - 2y - 2 = 0

    Simplifying:

    xy=1x - y = 1

  • Vertical Tangent Line occurs when dydx\frac{dy}{dx} is undefined, i.e., when the denominator equals zero:

    2x2y+6=02x - 2y + 6 = 0

    Simplifying:

    xy=3x - y = -3

Step 5: Solve for Points of Intersection

Now, we need to find the points where these lines intersect the original equation x22xy+y2+2x6y+21=0x^2 - 2xy + y^2 + 2x - 6y + 21 = 0.

  1. For horizontal tangents (xy=1x - y = 1):

    Substitute y=x1y = x - 1 into the original equation:

    x22x(x1)+(x1)2+2x6(x1)+21=0x^2 - 2x(x - 1) + (x - 1)^2 + 2x - 6(x - 1) + 21 = 0

    Simplifying this:

    x22x2+2x+x22x+1+2x6x+6+21=0x^2 - 2x^2 + 2x + x^2 - 2x + 1 + 2x - 6x + 6 + 21 = 0

    0x26x+28=00x^2 - 6x + 28 = 0

    Solving for xx:

    6x+28=0-6x + 28 = 0 x=286=143x = \frac{28}{6} = \frac{14}{3}

    Substituting x=143x = \frac{14}{3} into y=x1y = x - 1:

    y=1431=14333=113y = \frac{14}{3} - 1 = \frac{14}{3} - \frac{3}{3} = \frac{11}{3}

    So the point where the parabola has a horizontal tangent is (143,113)\left( \frac{14}{3}, \frac{11}{3} \right).

  2. For vertical tangents (xy=3x - y = -3):

    Substitute y=x+3y = x + 3 into the original equation:

    x22x(x+3)+(x+3)2+2x6(x+3)+21=0x^2 - 2x(x + 3) + (x + 3)^2 + 2x - 6(x + 3) + 21 = 0

    Simplifying this:

    x22x26x+x2+6x+9+2x6x18+21=0x^2 - 2x^2 - 6x + x^2 + 6x + 9 + 2x - 6x - 18 + 21 = 0

    0x2+12=00x^2 + 12 = 0

    This leads to a contradiction, so there are no solutions for vertical tangents.

Final Answer:

  • The parabola has horizontal tangent lines at the point (143,113)\left( \frac{14}{3}, \frac{11}{3} \right).
  • The parabola has no vertical tangent lines.

Would you like further clarification or additional questions on implicit differentiation? Here are some related questions:

  1. How does the chain rule affect implicit differentiation?
  2. What are the conditions

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Math Problem Analysis

Mathematical Concepts

Implicit Differentiation
Parabolas
Tangent Lines
Differential Calculus

Formulas

Implicit differentiation formula
Horizontal tangent condition (dy/dx = 0)
Vertical tangent condition (dy/dx undefined)

Theorems

Chain rule
Product rule
Quotient rule

Suitable Grade Level

Grades 11-12