You have y' = (2x-y)/(x-2y)
at x = -2, y = -1±√13, so there will be two lines tangent to the ellipse at x = -2.
for y = -1+√13,
y' = (-4+1-√13)/(-2-2(-1+√13))
= (-3-√13)/(-2√13)
= (13+3√13)/26
So, the line is
y = (13+3√13)/26 (x+2) + -1+√13
You can see the plots at
http://www.wolframalpha.com/input/?i=plot+x^2-xy%2By^2%3D16%2C+y+%3D+%2813%2B3%E2%88%9A13%29%2F26+%28x%2B2%29+%2B+%28-1%2B%E2%88%9A13%29
Now do the same for y = -1-√13
Find an equation of all tangent lines for the curve x^2-xy+y^2=16 x=-2
2x-(xy'+y(1))+2yy'=0
2x-xy'+y+2yy'=0
y'(-x+2y)=-2x-y
y'=-2x-y/-x+2y
y'=-2(0)-y/-(0)+2y
y'=-y/2y
Not sure what to do now?
1 answer