退役OIer只能学习文化课了qwq
文化课依然那么菜
只能一些自己不是很会的东西了
(图就自己画吧qwq)
S △ P F 1 F 2 = b 2 tan θ 2 θ = ∠ F 1 P F 2 \Large {S_{\triangle PF_1F_2 = b^2 \tan_{\frac{\theta}{2}}}} \quad \small \theta = ∠F_1PF_2 S△PF1F2=b2tan2θθ=∠F1PF2
证明
令 ∣ P F 1 ∣ = m , ∣ P F 2 ∣ = n |PF_1|=m,|PF_2|=n ∣PF1∣=m,∣PF2∣=n
∵ ∣ F 1 F 2 ∣ 2 = m 2 + n 2 − 2 m n cos θ \large{ \because |F_1F_2|^2 = m^2+n^2-2mn\cos {\theta}} ∵∣F1F2∣2=m2+n2−2mncosθ
∴ ∣ F 1 F 2 ∣ 2 = ( m + n ) 2 − 2 m n ( 1 + cos θ ) \large{ \therefore |F_1F_2|^2 = (m+n)^2 - 2mn(1+\cos \theta)} ∴∣F1F2∣2=(m+n)2−2mn(1+cosθ)
∴ 4 c 2 = 4 a 2 − 2 m n cos θ \large{\therefore 4c^2 = 4a^2-2mn \cos \theta} ∴4c2=4a2−2mncosθ
∴ m n = 2 b 2 1 + cos θ \large{\therefore mn = \frac{2b^2}{1+\cos \theta}} ∴mn=1+cosθ2b2
∴ S △ A B C = 1 2 m n sin θ = b 2 sin θ 1 + cos θ = b 2 tan θ 2 \large{\therefore S_{\triangle ABC}= \frac{1}{2}mn \sin \theta = \frac{b^2\sin \theta}{1+\cos \theta}= b^2 \tan{\frac{\theta}{2}}} ∴S△ABC=21mnsinθ=1+cosθb2sinθ=b2tan2θ
∣ M F 1 ∣ = a + e x 0 , ∣ M F 2 ∣ = a − e x 0 |MF_1| = a + ex_0, |MF_2| = a - ex_0 ∣MF1∣=a+ex0,∣MF2∣=a−ex0
证明
过点 M M M向准线 a 2 c \frac{a^2}{c} ca2做垂线,垂足为 N N N
∵ ∣ M F 1 ∣ ∣ M N ∣ = e \large{ \because \frac{|MF_1|}{|MN|} = e } ∵∣MN∣∣MF1∣=e
∴ ∣ M F 1 ∣ = e ∣ M N ∣ = c a × ( a 2 c − x 0 ) = a − e x 0 \large{\therefore |MF_1| = e|MN| = \frac{c}{a} \times (\frac{a^2}{c}-x_0) = a - ex_0} ∴∣MF1∣=e∣MN∣=ac×(ca2−x0)=a−ex0
d = ( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 = 1 + k 2 ∣ x 1 − x 2 ∣ = 1 + 1 k 2 ∣ y 1 − y 2 ∣ \large d = \sqrt{(x_1-x_2)^2 +(y_1-y_2)^2} = \sqrt{1+k^2}|x_1-x_2|=\sqrt{1+\frac{1}{k^2}}|y_1-y_2| d=(x1−x2)2+(y1−y2)2=1+k2∣x1−x2∣=1+k21∣y1−y2∣
d = e p ( 1 − e cos θ ) 2 P 为 焦 点 到 对 应 准 线 的 距 离 \large d = \frac{ep}{(1-e\cos \theta)^2} \quad \small P为焦点到对应准线的距离 d=(1−ecosθ)2epP为焦点到对应准线的距离
若点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)在椭圆 x 2 a 2 + y 2 b 2 = 1 \large {\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1} a2x2+b2y2=1上,则过点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)的切线方程为 x 0 x a 2 + y 0 y b 2 = 1 \large{\frac{x_0x}{a^2}+\frac{y_0y}{b^2}=1} a2x0x+b2y0y=1
若点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)在椭圆 x 2 a 2 + y 2 b 2 = 1 \large{\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1} a2x2+b2y2=1外,则过点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)做椭圆的两条切线切点为 P 1 , P 2 P1,P_2 P1,P2则弦 P 1 P 2 P_1P_2 P1P2的直线方程为 x 0 x a 2 + y 0 y b 2 = 1 \large{\frac{x_0x}{a^2}+\frac{y_0y}{b^2}= 1} a2x0x+b2y0y=1
证明略
证明基本同椭圆
S △ P F 1 F 2 = b 2 tan θ 2 \large S_{\triangle PF_1F_2} = \frac{b^2}{\tan \frac{\theta}{2}} S△PF1F2=tan2θb2
∣ P F 1 ∣ = e x 0 + a , ∣ P F 2 ∣ = e x 0 − a |PF_1| = ex_0+a,|PF_2| = ex_0-a ∣PF1∣=ex0+a,∣PF2∣=ex0−a
若点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)在双曲线 x 2 a 2 − y 2 b 2 = 1 \large \frac{x^2}{a^2}- \frac{y^2}{b^2}=1 a2x2−b2y2=1上,过点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)作切线方程为 x x 0 a 2 − y y 0 b 2 = 1 \large \frac{xx_0}{a^2}- \frac{yy_0}{b^2}=1 a2xx0−b2yy0=1
若点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)在双曲线 x 2 a 2 − y 2 b 2 = 1 \large \frac{x^2}{a^2}- \frac{y^2}{b^2}=1 a2x2−b2y2=1外,过点 P ( x 0 , y 0 ) P(x_0,y_0) P(x0,y0)作两条切线,切点分别为 P 1 P 2 P_1P_2 P1P2,则方程为 x x 0 a 2 − y y 0 b 2 = 1 \large \frac{xx_0}{a^2}- \frac{yy_0}{b^2}=1 a2xx0−b2yy0=1
1.点 P P P处的切线 P T PT PT平分 ∠ P F 1 F 2 ∠PF1F2 ∠PF1F2在点 P P P处的内角.