consider the parabola defined by the equation\n\n$x^{2}-2xy + y^{2}-2x - 2y + 17 = 0$.\n\nuse implicit…

consider the parabola defined by the equation\n\n$x^{2}-2xy + y^{2}-2x - 2y + 17 = 0$.\n\nuse implicit differentiation to find where the graph has horizontal and vertical tangent lines.\n\na. the parabola has horizontal tangent lines at the point(s)\n\nb. the parabola has vertical tangent lines at the point(s)

consider the parabola defined by the equation\n\n$x^{2}-2xy + y^{2}-2x - 2y + 17 = 0$.\n\nuse implicit differentiation to find where the graph has horizontal and vertical tangent lines.\n\na. the parabola has horizontal tangent lines at the point(s)\n\nb. the parabola has vertical tangent lines at the point(s)

Answer

Explanation:

Step1: Differentiate implicitly

Differentiate (x^{2}-2xy + y^{2}-2x - 2y+17 = 0) with respect to (x). Using the sum - rule ((u + v+w+\cdots)'=u'+v'+w'+\cdots), product - rule ((uv)' = u'v+uv') and chain - rule ((y^{n})'=ny^{n - 1}y'): [ \begin{align*} \frac{d}{dx}(x^{2})-\frac{d}{dx}(2xy)+\frac{d}{dx}(y^{2})-\frac{d}{dx}(2x)-\frac{d}{dx}(2y)+\frac{d}{dx}(17)&=0\ 2x-(2y + 2xy')+2yy'-2 - 2y'&=0 \end{align*} ]

Step2: Solve for (y')

[ \begin{align*} 2x-2y-2xy'+2yy'-2 - 2y'&=0\ y'(-2x + 2y-2)&=2y - 2x+2\ y'&=\frac{2y - 2x + 2}{-2x + 2y-2}=\frac{y - x + 1}{-x + y-1} \end{align*} ]

Step3: Find horizontal tangent lines ((y'=0))

Set (y' = 0), then (y-x + 1=0), so (y=x - 1). Substitute (y=x - 1) into the original equation (x^{2}-2x(x - 1)+(x - 1)^{2}-2x-2(x - 1)+17 = 0) [ \begin{align*} x^{2}-2x^{2}+2x+x^{2}-2x + 1-2x-2x + 2+17&=0\ -4x+20&=0\ x&=5 \end{align*} ] When (x = 5), (y=x - 1=4)

Step4: Find vertical tangent lines (denominator of (y'=0))

Set (-x + y-1=0), so (y=x + 1). Substitute (y=x + 1) into the original equation (x^{2}-2x(x + 1)+(x + 1)^{2}-2x-2(x + 1)+17 = 0) [ \begin{align*} x^{2}-2x^{2}-2x+x^{2}+2x + 1-2x-2x - 2+17&=0\ -4x + 16&=0\ x&=4 \end{align*} ] When (x = 4), (y=x + 1=5)

Answer:

a. ((5,4)) b. ((4,5))