# Thread: Horizontal Tangent - Implicit Differentiation

1. ## Horizontal Tangent - Implicit Differentiation

a curve in the plane is given by the following equation
x^4+2x^2y+3y^2=6

A) Find an expression for dy/dx using implicit differentiation
B) Find the four points on the curve where the tangent line is horizontal

Ok I know how to solve for the derivative. I got
dy/dx = (4x^3-4xy)/(2x^2+6y)

However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.

2. Originally Posted by rawkstar
a curve in the plane is given by the following equation
x^4+2x^2y+3y^2=6

A) Find an expression for dy/dx using implicit differentiation
B) Find the four points on the curve where the tangent line is horizontal

Ok I know how to solve for the derivative. I got
dy/dx = (4x^3-4xy)/(2x^2+6y)

However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.
I am assuming your derivative is correct ...

$\frac{dy}{dx} = 0$ if $4x^3 - 4xy = 0$

$4x(x^2-y) = 0$

$x = 0$ , $y = x^2$

sub $x = 0$ and find the corresponding y-value(s)

sub $y = x^2$ and find the corresponding x-values

3. sub it into the original equation?

4. Originally Posted by rawkstar
sub it into the original equation?
yes ... you are trying to find points on the curve.

5. ok so i subbed those into the orignial equation and got two x values and two y values
do i then plug those into the original to get the corresponding opposite value

6. Originally Posted by rawkstar
ok so i subbed those into the orignial equation and got two x values and two y values
do i then plug those into the original to get the corresponding opposite value
when you sub in x = 0 , you get two values for y ... the two points will be $(0,\sqrt{2})$ and $(0,-\sqrt{2})
$

when you sub in $x^2$ for $y$, you get two values for y ... square those to get the corresponding x-values.