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)

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:

[tex]\frac{dy}{dx}=0[/tex] (The slope is zero)

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

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

Derivating respect to x:

[tex]\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}[/tex]=[tex]2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0[/tex]

Solving for dy/dx:

[tex]\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}[/tex]

Applying the first conditon (slope is zero)

[tex]\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0[/tex]

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

y=x+2 (I)

Replacing (I) in the given equation:

[tex]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[/tex]

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:

[tex]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[/tex]

Replacing it in (II)

y=1+4

y=5

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

batolisis batolisis · Answer 2 · 2021-10-26T12:31:19+02:00

Answer:

The parabola has a horizontal tangent line at the point [tex](2,4)[/tex]

The parabola has a vertical tangent line at the point [tex](1,5)[/tex]

Step by Step Explanation:

When the equation [tex]x^{2} -2xy+y^{2}+4x-8y+20=0[/tex] is implicitly differentiated, we get

[tex]y'=\frac{x-y+2}{x-y+4}[/tex]

(1) For horizontal tangent line, [tex]y'=0[/tex]

[tex]\frac{x-y+2}{x-y+4}=0[/tex]

[tex]{x-y+2}=0[/tex]

[tex]y=x+2[/tex]

To find the tangent point, substitute [tex]y=x+2[/tex] into [tex]x^{2} -2xy+y^{2}+4x-8y+20=0[/tex], and solve for x

[tex]x^{2} -2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0[/tex]

Clear brackets and solve for x, this gives

[tex]x=2[/tex]

Substitute [tex]x=2[/tex] into the equation [tex]y=x+2[/tex] to find y

[tex]y=x+2[/tex]

[tex]y=(2)+2=4[/tex]

The parabola has a horizontal tangent point at [tex](x, y)=(2,4)[/tex]

(1) For vertical tangent line, y' is undefined

This means that the denominator of y' is zero

[tex]{x-y+4}=0[/tex]

[tex]y=x+4[/tex]

To find the tangent point, substitute [tex]y=x+4[/tex] into [tex]x^{2} -2xy+y^{2}+4x-8y+20=0[/tex], and solve for x

[tex]x^{2} -2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0[/tex]

Clear brackets and solve for x, this gives

[tex]x=1[/tex]

Substitute [tex]x=1[/tex] into the equation [tex]y=x+4[/tex] to find y

[tex]y=x+4[/tex]

[tex]y=(1)+4=5[/tex]

The parabola has a vertical tangent point at [tex](x, y)=(1,5)[/tex]

For any curve,

(1) A horizontal tangent line occurs when y' is zero.

(2) A vertical tangent line occurs when y' is undefined

(that is when the denominator is zero)

Learn more about horizontal and vertical tangents here

https://brainly.com/question/2141830