tangent x^2-xy-y^2=1,(2,1)

解答

切线

x^{2} - x y - y^{2} = 1, (2, 1)

y = \frac{3}{4} x - \frac{1}{2}

求解步骤

计算

x^{2} - x y - y^{2} = 1

的斜率:

\frac{d y}{d x} = - \frac{- 2 x + y}{x + 2 y}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{3}{4}

找到斜率 m=

\frac{3}{4}

且穿过

(2, 1)

的直线:

y = \frac{3}{4} x - \frac{1}{2}

y = \frac{3}{4} x - \frac{1}{2}

\int x^{2} + 2 x + 1 d x

\frac{d y}{d x} = (\frac{x y + 5 x - y - 5}{x y - 2 x + 6 y - 12})

x y^{3} d x + e^{x^{2}} d y = 0

(5 x^{2} + 2 x^{- 1} y + 2 x^{- 1} y^{2}) d x + (1 + 2 y) d y = 0

d er i v a t i v e e^{θ} sin (5 θ)