How do you find the horizontal and vertical tangents of #4x^2 + y^2 -8x +4y +4=0#?

Answer 1

Horizontal tangents at: #(1,0)# and #(1,-4)#
Vertical tangents at: #(0,-2)# and #(2,-2)#

We have:

# 4x^2 + y^2 -8x +4y +4=0 #

Differentiating implicitly:

# 8x + 2ydy/dx -8 +4dy/dx = 0 # # :. (y+2)dy/dx =4-4x # # :. dy/dx =(4-4x)/(y+2) #
We have horizontal tangents when #dy/dx=0#
# => 4-4x = 0 # # :. \ \ x = 1 #
With #x=1# we have:
# 4 + y^2 -8 +4y +4=0 # # :. y^2 +4y =0 # # :. y(y+4) =0 # # :. y=0,-4#
ie the coordinates #(1,0)# and #(1,-4)#
We have vertical tangents when #dy/dx rarr oo# (Note I did not write #dy/dx = oo# as a mathematician I really dislike this abuse of notation, #oo# is not a number, and cannot be equated).
# => y+2 = 0 # # :. \ \ y = -2 #
With #y=-2# we have:
# 4x^2 + 4 -8x -8 +4=0 # # :. 4x^2 -8x = 0 # # :. 4x(x-2) = 0 # # :. x=0,2 #
ie the coordinates #(0,-2)# and #(2,-2)#

We can confirm this by completing the square:

# 4x^2 -8x +y^2 + 4y +4=0 # # 4{x^2 -2x} +{y^2 + 4y} +4=0 # # 4{(x-1)^2-1} +{(y+2)^2 - 4} +4=0 # # 4(x-1)^2 -4+(y+2)^2 -4+4=0 # # 4(x-1)^2 +(y+2)^2 =4 # # (x-1)^2 +(y+2)^2/4 =1 # # (x-1)^2 +((y+2)/2)^2 =1 #
An ellipse of centre #(1,-2)# with minor and major axis #1# and #2# respectively.

graph{4x^2 + y^2 -8x +4y +4=0 [-1, 3, -6, 2]}

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Answer 2

To find the horizontal and vertical tangents of the equation 4x^2 + y^2 -8x +4y +4=0, we can differentiate the equation with respect to x and y separately.

Differentiating with respect to x, we get: 8x - 8 + 0 + 0 = 0 Simplifying, we find: x = 1

Differentiating with respect to y, we get: 0 + 2y + 0 + 4 = 0 Simplifying, we find: y = -2

Therefore, the horizontal tangent occurs at the point (1, -2) and the vertical tangent occurs at the same point (1, -2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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