Find all points on the curve 2xy^2 - 25x^2 y =5 where it has a horizontal tangent line.

Question

Write each point as a coordinate pair, and separate multiple points with a comma.

Answer 1

Let F (x , y) = 2 x y² - 25 x² y where y = h ( x )

Then d y / d x = - F' _x (x , y) / F '_y (x , y) = [ 50 x y - 2 y² ] / [ 4 x y - 25 x² ]

d y / d x = 0 ⇔ 50 x y + 2 y² = 0 and since can not be equal to 0 ,then y = 25 x .

Substituting y = - 25 x back to 2 x y² - 25 x² y = 5 we get x³ = 1 / 125 That is x = 1 / 5 and therefor y = 5.

Then we can conclude that the only horizontal tangent line happens at the point ( 1 / 5 , 5 )

Answer 2

Horizontal tangent lines occur at all points when dy/dx is 0.

So we'll solve for dy/dx then set that equal to 0

Use the product rule

2x * (2y dy/dx) + y² * (2) - 25x² * (dy/dx) + y* (-50x) = 0

Instead of solving for dy/dx, we can actually just plug in 0 for it now

2x * 0 + 2y² - 25x² * 0 -50xy = 0

2y² - 50xy = 0

2y(y-25x) = 0

We have a horizontal tangent line when either 2y = 0 or y -25x = 0

Therefore, we have a horizontal tangent line when y = 0 or y = 25x

These however, are not points on our curve.

To get the points, we can plug 0 in for y in our equation and 25x in for y in our equation.

Plug in 0 first:

2x(0)² - 25x²(0) = 5

0 = 5

This makes no sense whatsoever so we have no horizontal tangent lines when y = 0.

Now let's plug in 25x:

2x(25x)² - 25x²(25x) = 5

1250x³ - 625x³ = 5

625x³ = 5

x³ = 5/625 = 1/125

x = 1/5

Solve for y now.

y = 25(1/5) = 5

Therefore, we have a horizontal tangent line at (1/5, 5)