Asked by Boan
Which points on the curve of x^2 - xy - y^2 = 5 have vertical tangent lines?
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Answered by
Anonymous
where is dy/dx = + or - oo ????
2 x dx - x dy - y dx - 2 y dy = 0
(2x-y) dx - (2y+x)dy = 0
dy/dx = (2x-y)/(2y+x)
when is 2y+x = 0?
when x = -2y
(-2y)^2 + 2 y^2 -y^2 = 5
3 y^2 = 5
y = +/- sqrt (5/3)
x - -/+ 2 sqrt(5/3)
check my arithmetic. I did that fast.
2 x dx - x dy - y dx - 2 y dy = 0
(2x-y) dx - (2y+x)dy = 0
dy/dx = (2x-y)/(2y+x)
when is 2y+x = 0?
when x = -2y
(-2y)^2 + 2 y^2 -y^2 = 5
3 y^2 = 5
y = +/- sqrt (5/3)
x - -/+ 2 sqrt(5/3)
check my arithmetic. I did that fast.
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