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# Finding coordinates on curve where gradient of normal is 1

+1
186
4

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

Mar 25, 2019
edited by bqrs01  Mar 25, 2019
edited by bqrs01  May 14, 2019

#1
+4

Perhaps this graph will help: Mar 25, 2019
#2
+1

I have a further question, though. How would I determine this other coordinate if I didn't have access to a graph?

Is it possible though algebra alone? If so, could you please show me how?

bqrs01

bqrs01  Mar 25, 2019
edited by bqrs01  May 14, 2019
#3
+1

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 )   Mar 25, 2019
edited by CPhill  Mar 25, 2019
#4
+2

I was trying to solve the following question: $$\begin{array}{|rcll|} \hline x^2(x+3y)-y^3 &=& 3 \\ x^3 + 3x^2y-y^3 - 3 &=& 0 \\\\ \mathbf{f(x,y)} & \mathbf{=} & \mathbf{x^3 + 3x^2y-y^3 - 3} \\\\ \dfrac{\partial f} {\partial x} &=& 3x^2+6xy \\ \dfrac{\partial f} {\partial y} &=& 3x^2-3y^2 \\ && \boxed{\text{Formula: }\\ \dfrac{dy}{dx} = -\dfrac{\dfrac{\partial f} {\partial x}} {\dfrac{\partial f} {\partial y}} } \\ \dfrac{dy}{dx} &=& -\dfrac{ 3x^2+6xy } { 3x^2-3y^2 } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ 3(x^2+2xy) } { 3(x^2-y^2) } \\\\ \dfrac{dy}{dx} &=& -\dfrac{ x^2+2xy } { x^2-y^2 } \\\\ \mathbf{ \dfrac{dy}{dx}} &\mathbf{=}& \mathbf{\dfrac{ x^2+2xy } { y^2-x^2 }} \\ \hline \end{array}$$ Mar 26, 2019