Find the equations of the tangent and normal to the following curves at the point indicated.

x²-2xy-2y² = 1; (-3, -1)

Help please :)

Question

Find the equations of the tangent and normal to the following curves at the point indicated.

x²-2xy-2y² = 1; (-3, -1)

Help please :)

anonymous · Answer

i think its like this.... not so sure uh...

since y' = slope = m

find y' :
2x-2(xy'+y)=0
2x-2xy'-2y=0
x-xy'-y=0
y'=(x-y)/x

for tangent :
y+1=(x-y)/x  * (x+3)

for normal:
y+1=-x/(x-y)   * (x+3)

anonymous · Answer

Alright, thanks. I'm gonna figure it too. :)

nikvist · Answer

$$x^2-2xy-2y^2=1$$$$(x^2-2xy-2y^2)'=0$$$$2x-2y-2xy'-4yy'=0$$$$x-y=(x+2y)y'$$$$y'=\frac{x-y}{x+2y}$$$$y'(-3,-1)=\frac{-3-(-1)}{-3+2(-1)}=\frac{-2}{-5}=\frac{2}{5}$$ tangent: $$y+1=\frac{2}{5}(x+3)$$ normal: $$y+1=-\frac{5}{2}(x+3)$$