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    Any chance anybody can help me out with this, I think it's straight forward enough I must be doing something silly;

    A curve has the equation

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

    Find the co-ordinates at the point on the curve where \frac{dy}{dx}=0

    When I differentiated I got
     2x + 2y + 2x\frac{dy}{dx} -6y\frac{dy}{dx}=0

    Which I then rearranged as;

    \frac{dy}{dx}= \frac{-2x - 2y}{2x-6y}

    which I then put =0 and got confused.

    Is what I've done so far right? and how do I finish it?

    Thankyou
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    (Original post by Becky7337)
    Any chance anybody can help me out with this, I think it's straight forward enough I must be doing something silly;

    A curve has the equation

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

    Find the co-ordinates at the point on the curve where \frac{dy}{dx}=0

    When I differentiated I got
     2x + 2y + 2x\frac{dy}{dx} -6y\frac{dy}{dx}=0

    Which I then rearranged as;

    \frac{dy}{dx}= \frac{-2x - 2y}{2x-6y}

    which I then put =0 and got confused.

    Is what I've done so far right? and how do I finish it?

    Thankyou
    Your differentiation was right. You need to set that equal to 0 and solve by substituting for x or y.
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    If A/B = 0, what can you say about A?
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    Yeah i think what you've done so far is correct. You put it equal to 0, which means you can multiply the bottom to get rid of it, leaving you with an equation with x and y equal to 0. You can then rearrange to make either x or y the subject and sub it back in, to find either the X or Y coordinate
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    (For future reference) Don't bother to rearrange, substitute dy/dx = 0 into 2x + 2y + 2x \dfrac{\text{d}y}{\text{d}x} - 6y \dfrac{\text{d}y}{\text{d}x} = 0. Therefore 2x + 2y = 0 and x = -y, which can be substituted into the equation of the curve.
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    ok thankyou...you guys are really quick..I only turned my head for a second!

    I get it now I was just a little confused when I got x + y=0

    Thankyou
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    Ie the numerator must be zero as the denominator cannot vanish.
 
 
 
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