Points where gradient is zero (plotting it)

[W3bbo]

Homework Statement

A curve has equation:

x^2+2xy-3y^2+16=0

Find the co-ordinates of the points on the curve where dy/dx=0

I think I was able to differentiate it and get the coordinates fine, but I'm wanting to plot the function in Mathematica (5.2) to see if I'm right or not (BTW, I tried Ma's Dt[] and Differential[] functions, but I can't interpret the results. And plot[f, {x,-2,2}] just gives me error messages because y is undefined).

2. The attempt at a solution

x^2+2xy-3y^2+16=0

2x+2x(dy/dx)+y-3(2y(dy/dx))=0

y+2x+(dy/dx)(2x-6y)=0

(dy/dx)=-(y+2x)/(2x-6y)=0

For the fraction to equal zero, the numerator must also be zero, therefore:

-y-2x=0
y=-2x

Given this, substituting this value for y:

x^2+2x(-2x)-3(-2x)^2+16=0
x^2-4x^2-12x^2+16=0
-15x^2+16=0

Therefore (using the quadratic formula):

x=Sqrt(960)/-30
x=Sqrt(960)/30

but it seems a little hackish to me, this from a past-paper (Edexcel Advanced Level C4, 28th June 2005), usually you get integer answers.

But besides asking if I'm right, how can I plot functions with multiple instances of x and y within? I'm guessing I'd need to convert it to a parametric somehow.

 

[f(x)]

A curve has equation:

x^2+2xy-3y^2+16=0

2. The attempt at a solution

x^2+2xy-3y^2+16=0

2x+2x(dy/dx)+y-3(2y(dy/dx))=0

y+2x+(dy/dx)(2x-6y)=0

diff. gives \ 2x + 2x \frac{dy}{dx} + 2y - 3(2y \frac{dy}{dx} ) = 0

 

[W3bbo]

diff. gives \ 2x + 2x \frac{dy}{dx} + 2y - 3(2y \frac{dy}{dx} ) = 0

Where did +2y come from? I didn't have a solitary y^2 expression.

EDIT: Ah I see, product rule; I forgot to reapply the coefficient (2) of xy after performing the product differentiation.

Still, how can I plot the function?

 

[Mathgician]

the question is asking for the critical points of the surface right? I have a question do you need to graph this function? Do you need to find the saddle points and min max?

 

[W3bbo]

the question is asking for the critical points of the surface right? I have a question do you need to graph this function? Do you need to find the saddle points and min max?

I'm not being asked to plot the graph, and I've since found what I think are the right co-ordinates (by substituting the resolved value of y into the equation and solving) as { (2,0) , (-2,0) }.

I want to plot the graph out of personal curiosity, to see what the graph actually looks like (but also to make sure I'm right). I haven't covered the plotting of implicit functions on my curriculum's syllabus though. Hence why I'm asking :)