Horizontal and vertical tangents

[apbaldasare]

Alright, so this is one of the implicit differentiation problems. I had to find the dy/dx for a curve defined by y⁴ =y² -x² I got dy/dx to be x/(2y-y^3)

What would be the horizontal and vertical tangent to this curve? Thanks!

 

[galactus]

Check your y' again. I think it should be

\(\displaystyle y'=\frac{x}{y(2y^{2}-1)}\)

To find the y coordinate of the vertical tangents, set the denominator equal to 0 and solve for y.

The corresponding x coordinate can then be found by plugging that y back into the original equation.

Vertical tangents occur where the slope is 'infinite'. Thus, it is undefined. So, set the denominator equal to 0 and solve for y.

The horizontal tangents are where the slope is 0.

Set y' equal to 0 and x is easy to see.

 

[Deleted member 4993]

Alright, so this is one of the implicit differentiation problems. I had to find the dy/dx for a curve defined by y⁴ =y² -x² I got dy/dx to be x/(2y-y^3)

What would be the horizontal and vertical tangent to this curve? Thanks!

For vertical assymptote:

At what values of (x,y) y' becomes undefined?

For horizontal assymptote - rewrite your function as:

x² = (y - y²)(y + y²)

Do you see any range of y that would be forbidden for this expression?