Question

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=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)

Answer 1

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$=$2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$

Solving for dy/dx:

$\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}$

Applying the first conditon (slope is zero)

$\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0$

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

$x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2$

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

$x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1$

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)