tangent x^2+y^2=169,(-5,12)

解答

切线

x^{2} + y^{2} = 169, (- 5, 12)

y = \frac{5}{12} x + \frac{169}{12}

求解步骤

计算

x^{2} + y^{2} = 169

的斜率:

\frac{d y}{d x} = - \frac{x}{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{5}{12}

找到斜率 m=

\frac{5}{12}

且穿过

(- 5, 12)

的直线:

y = \frac{5}{12} x + \frac{169}{12}

y = \frac{5}{12} x + \frac{169}{12}

l a pl a ce t r an s f or m 1 - 2 t

x \to \infty lim (π x)

d er i v a t i v e 7 t + 2 t^{7}

\frac{d}{d x} (x f (x) + f (x) (c x) + c f (x))

\int_{1}^{2} - x + \frac{85}{50} d x