已知圆C:(x+4)^2+y^2=4,圆D的圆心D在y轴上且与圆C外切,圆D与Y轴交于A,B两点,且P(-3,0)(1)若点D(0,3),求∠APB的正切值;(2)当点D在y轴上运动时,求tan∠APB的最大值;
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![已知圆C:(x+4)^2+y^2=4,圆D的圆心D在y轴上且与圆C外切,圆D与Y轴交于A,B两点,且P(-3,0)(1)若点D(0,3),求∠APB的正切值;(2)当点D在y轴上运动时,求tan∠APB的最大值;](/uploads/image/z/15250822-70-2.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86C%EF%BC%9A%28x%2B4%29%5E2%2By%5E2%3D4%2C%E5%9C%86D%E7%9A%84%E5%9C%86%E5%BF%83D%E5%9C%A8y%E8%BD%B4%E4%B8%8A%E4%B8%94%E4%B8%8E%E5%9C%86C%E5%A4%96%E5%88%87%2C%E5%9C%86D%E4%B8%8EY%E8%BD%B4%E4%BA%A4%E4%BA%8EA%2CB%E4%B8%A4%E7%82%B9%2C%E4%B8%94P%28-3%2C0%29%EF%BC%881%EF%BC%89%E8%8B%A5%E7%82%B9D%EF%BC%880%2C3%EF%BC%89%2C%E6%B1%82%E2%88%A0APB%E7%9A%84%E6%AD%A3%E5%88%87%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E5%BD%93%E7%82%B9D%E5%9C%A8y%E8%BD%B4%E4%B8%8A%E8%BF%90%E5%8A%A8%E6%97%B6%2C%E6%B1%82tan%E2%88%A0APB%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%EF%BC%9B)
已知圆C:(x+4)^2+y^2=4,圆D的圆心D在y轴上且与圆C外切,圆D与Y轴交于A,B两点,且P(-3,0)(1)若点D(0,3),求∠APB的正切值;(2)当点D在y轴上运动时,求tan∠APB的最大值;
已知圆C:(x+4)^2+y^2=4,圆D的圆心D在y轴上且与圆C外切,圆D与Y轴交于A,B两点,且P(-3,0)
(1)若点D(0,3),求∠APB的正切值;
(2)当点D在y轴上运动时,求tan∠APB的最大值;
已知圆C:(x+4)^2+y^2=4,圆D的圆心D在y轴上且与圆C外切,圆D与Y轴交于A,B两点,且P(-3,0)(1)若点D(0,3),求∠APB的正切值;(2)当点D在y轴上运动时,求tan∠APB的最大值;
(1)
C(-4,0),圆C的半径为r = 2
CD² = (-4 - 0)² + (0 - 3)² = 25,CD = 5,圆D的半径为R = CD - r = 5 - 2 = 3
不妨设A上B下,A(0,6),B(0,0)
tan∠APB = AB/PB = 6/3 = 2
(2)
D(0,d)
R = √(16 + d²) - 2
A(0,d + R),B(0,d - R)
PA的斜率u = (d + R)/3
PB的斜率v = (d - R)/3
f(d) = tan∠APB = (u - v)/(1 + uv)
= 6R/(9 + d² - R²)
= 6[√(16 + d²) - 2]/[4√(16 + d²) - 11]
f'(d) = -18d/{√(16 + d²)[4√(16 + d²) - 11]²}= 0
d = 0
tan∠APB的最大值 = 6[√(16 + 0) - 2]/[4√(16 + 0) - 11]
= 12/5