tangent x^2+xy+2y^2=16,(-2,-2)

解答

切线

x^{2} + x y + 2 y^{2} = 16, (- 2, - 2)

y = - \frac{3}{5} x - \frac{16}{5}

求解步骤

计算

x^{2} + x y + 2 y^{2} = 16

的斜率:

\frac{d y}{d x} = \frac{- 2 x - y}{x + 4 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}{5}

找到斜率 m=

- \frac{3}{5}

且穿过

(- 2, - 2)

的直线:

y = - \frac{3}{5} x - \frac{16}{5}

y = - \frac{3}{5} x - \frac{16}{5}

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