Question

horizontal tangent and vertical tangent
I need to find the coordinates of the points where the curve has a horizontal and vertical tangent

Answer 1

\[4x^2+y^2-8x+4y+4-0\]

Answer 2

For horizontal tangents dy/dx = 0

For vertical tangents dy/dx = \(\infty\)

Answer 3

I took the derivative and for the horizontal tangent, I set dy/dx =0 and solve for it, I got (1,0) and (1,-4)

Answer 4

yesterday someone told me that dx/dy=0 is how I find the vertical tangent

Answer 5

No that's for horizontal tangents

Answer 6

You are absolutely correct for the points for horizontal tangents.

Answer 7

I got (2, -2) and (0,-2)

Answer 8

Correct.

Answer 9

for the vertical tangent

Answer 10

but I did set up dy/dx=0 and solve for x then sub those values back into the original for the horizontal tangent

For the vertical tangent I did set up dx/dy=0 and solve for y and then sub back into the original

I also, took the given curve, and wrote it in standard form for the ellipse to double check my solutions

Answer 11

You are correct.

Instead of putting dy/dx or dx/dy = 0, you can directly calculate :

\[\large{\cfrac{dy}{dx} = \cfrac{4-4x}{y+2}}\]

And then find out the values of x and y for dy/dx to be 0 or \(\infty\)

Answer 12

Thanks, I did not see the quicker method to doing these types of problems.

Answer 13

:) No problem.