+53 Hi there.
I was trying to solve the following question:
I completed part (i), by showing that dy/dx is equal to the stated above.
However, when I attempted part (ii), I got only one coordinate (1, -2), (I equated dy/dx = -1).
How would I go about finding the other coordinate?
Thank you so much in advance,
bqrs01
+129849 x^2 + 2xy
________ = -1
y^2 - x^2
x^2 + 2xy = -1 ( y^2 - x^2)
x^2 + 2xy = x^2 - y^2
2xy = -y^2
y^2 + 2xy = 0
y ( y + 2x) = 0 set both factors to 0
y = 0 and subbing this into the original equation produces
x^3 = 3
x= (3)^(1/3)
So...another point is ( 3^(1/3) , 0 )
+26387 I was trying to solve the following question:
\(\begin{array}{|rcll|} \hline x^2(x+3y)-y^3 &=& 3 \\ x^3 + 3x^2y-y^3 - 3 &=& 0 \\\\ \mathbf{f(x,y)} & \mathbf{=} & \mathbf{x^3 + 3x^2y-y^3 - 3} \\\\ \dfrac{\partial f} {\partial x} &=& 3x^2+6xy \\ \dfrac{\partial f} {\partial y} &=& 3x^2-3y^2 \\ && \boxed{\text{Formula: }\\ \dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f} {\partial x}} {\dfrac{\partial f} {\partial y}} } \\ \dfrac{dy}{dx} &=& -\dfrac{ 3x^2+6xy } { 3x^2-3y^2 } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ 3(x^2+2xy) } { 3(x^2-y^2) } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ x^2+2xy } { x^2-y^2 } \\\\ \mathbf{ \dfrac{dy}{dx}} &\mathbf{=}& \mathbf{\dfrac{ x^2+2xy } { y^2-x^2 }} \\ \hline \end{array}\)
