The question states: the graph of x^2+4y^2-4x-12y+4=0 has two points of horizontal tangency, find those points.
I think I found dy/dx correctly. I got dy/dx=(4-2x)/(8y-12)=(2-x)/(4y-6), but that may be wrong. I just don't know where to go from there. Any help is appreciated.
I think I found dy/dx correctly. I got dy/dx=(4-2x)/(8y-12)=(2-x)/(4y-6), but that may be wrong. I just don't know where to go from there. Any help is appreciated.
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Implicitly differentiate:
2x dx + 8y dy - 4dx -12dy=0
dy(8y-12) + dx(2x-4)=0
dy(8y-12)=dx(4-2x)
dy/dx=(4-2x)/(8y-12)
now dy/dx=0 when x=2
Now we plug 2 in for x in original equation: 4+4y^2-8-12y+4=0
4y^2-12y=0 or y^2-3y=0 and y(y-3)=0 and y=[0,3]
So when x=2, y=[0,3] and our two tangent points are (2,0) and (2,3)
2x dx + 8y dy - 4dx -12dy=0
dy(8y-12) + dx(2x-4)=0
dy(8y-12)=dx(4-2x)
dy/dx=(4-2x)/(8y-12)
now dy/dx=0 when x=2
Now we plug 2 in for x in original equation: 4+4y^2-8-12y+4=0
4y^2-12y=0 or y^2-3y=0 and y(y-3)=0 and y=[0,3]
So when x=2, y=[0,3] and our two tangent points are (2,0) and (2,3)
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Remember that dy/dx is the same as the slope of the tangent. When the tangent is horizontal is means that the slope is equal to zero. So let dy/dx=0 and you should be able to get it out.