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

解答

切线

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

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

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{12}{5}

找到斜率 m=

\frac{12}{5}

且穿过

(- 12, 5)

的直线:

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

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

n = 6 \sum \infty \frac{7 ^{n}}{1 2 ^{n}}

\int_{0}^{3} 4 π (\frac{x ^{3}}{3}) d x

d er i v a t i v e 1 + 10 t

t an g e n t f (x) = \frac{x ^{3}}{4}, (- 6, - 54)

\frac{\partial}{\partial x} (sin (3 x) cos (2 y))