Chapter 4, Problem 74RE

a)

To determine

To determine slope of the parabola at each of the points where it crosses the coordinate axes vertically.

Answer to Problem 74RE

Slope of parabola where tangent line is vertical is infinity $\infty$

Explanation of Solution

Given: Equation of parabola $x^2-2xy+y^2-4x=8$

Concept Used: Differentiation with respect to $x$

Calculation: We have equation $x^2-2xy+y^2-4x=8$

Differentiate above equation with respect to $x$ we get,

  $2x-\left(2x\frac{dy}{dx}+2y\right)+2y\frac{dy}{dx}-4=0$

  $\frac{dy}{dx}\left(2y-2x\right)=2y-2x+4$

  $\frac{dy}{dx}=\frac{2y-2x+4}{2y-2x}=\frac{y-x+2}{y-x}$

Tangent will be vertical when denominator of the derivative equals to 0

  $y-x=0$

  $y=x$

Put $y=x$ in original equation, we get

  $x^2-2x\left(x\right)+{\left(x\right)}^2-4x=8$

  $-4x-8=0$

  $-4x=8$

  $x=-2$

Now put $x=-2$ in $y=x$ we get,

  $y=-2$

Hence, the point where the tangent line is vertical is $\left(-2,-2\right)$

Now we find the slope of parabola by putting point $\left(-2,-2\right)$ in $\frac{dy}{dx}$ we get,

  $\frac{dy}{dx}=\frac{2y-2x+4}{2y-2x}=\frac{y-x+2}{y-x}$

  $\frac{dy}{dx}=\frac{-2+2+2}{-2+2}=\frac{2}{0}=\infty$

Conclusion: Slope of parabola where tangent line is vertical is infinity $\infty$

b)

To determine

To determine slope of the parabola at each of the points where it crosses the coordinate axes horizontally.

Answer to Problem 74RE

Slope of parabola where tangent line is horizontal is $0$

Explanation of Solution

Given: Equation of parabola $x^2-2xy+y^2-4x=8$

Concept Used: Differentiation with respect to $x$

Calculation: We have equation $x^2-2xy+y^2-4x=8$

Differentiate above equation with respect to $x$

  $2x-\left(2x\frac{dy}{dx}+2y\right)+2y\frac{dy}{dx}-4=0$

  $\frac{dy}{dx}\left(2y-2x\right)=2y-2x+4$

  $\frac{dy}{dx}=\frac{2y-2x+4}{2y-2x}=\frac{y-x+2}{y-x}$

Tangent will be horizontal when numerator of $\frac{dy}{dx}=0$

  $y-x+2=0$

  $y=x-2$

Now, substitute the $y=x-2$ in $x^2-2xy+y^2-4x=8$ and we get,

  $x^2-2x(x-2)+{(x-2)}^2-4x=0$

  $x^2-2x^2+4x+x^2+4-4x-4x=8$

  $4-4x=8$

  $x-1$

Now put $x=-1$ in $y=x-2$ we get,

  $x=-1,y=-3$

Hence, the point where the tangent is horizontal is $\left(-1,-3\right)$

Now we find the slope of parabola by putting point $\left(-1,-3\right)$ in $\frac{dy}{dx}$ we get,

  $\frac{dy}{dx}=\frac{2y-2x+4}{2y-2x}=\frac{y-x+2}{y-x}$

  $\frac{dy}{dx}=\frac{-3+1+2}{-3+1}=\frac{-3+3}{-2}=0$

Conclusion: Slope of parabola where tangent line is horizontal is $0$