Chapter 4, Problem 73RE

a)

To determine

To determine the coordinates of point $A$ of given parabola where tangent line is vertical.

Answer to Problem 73RE

Point is $\left(-2,-2\right)$

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,

  $y=-2$

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

b)

To determine

To determine the coordinates of point $B$ of given parabola where tangent line is horizontal.

Answer to Problem 73RE

Point is $\left(-1,-3\right)$

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,

  $x=-1,y=-3$

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