关于圆锥曲线的数学题
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8-d²\u22640\uff0c
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8-d²\u22650\uff0c\u5f97
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2.\u8bbe\u629b\u7269\u7ebfy²=2x\u7684\u7126\u70b9\u4e3aF\uff0c\u8fc7\u70b9M(\u221a3,0)\u7684\u76f4\u7ebf\u4e0e\u629b\u7269\u7ebf\u76f8\u4ea4\u4e0eA\uff0cB\u4e24\u70b9\uff0c\u4e0e\u629b\u7269\u7ebf\u7684\u51c6\u7ebf\u76f8\u4ea4\u4e0eC\uff0c\uff5cBF\uff5c=2\uff0c\u5219\u25b3BCF\u4e0e\u25b3ACF\u6240\u6210\u7684\u9762\u79ef\u4e4b\u6bd4\u4e3a\uff1f
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y\u2032=k(x\u2032-\u221a3\uff09\uff0c\u2460
y\u2032²=2x\u2032\uff0c \u2461
y\u2032²+\uff08x\u2032-1/2)²=4\u3002\u2462
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\u628ax\u2032=3/2\u548cy\u2032=\u221a3\u4ee3\u5165\u2460\u5f97k=\u221a3/(3/2-\u221a3\uff09\uff0c
\u628ak=\u221a3/(3/2-\u221a3\uff09\u4ee3\u5165y=k(x-\u221a3),\u5f97y=(\u221a3x-3\uff09/(3/2-\u221a3\uff09\uff0c
\u76f4\u7ebfy=(\u221a3x-3\uff09/(3/2-\u221a3\uff09\u4e0e\u76f4\u7ebfx=-1/2\u7684\u4ea4\u70b9\u7684\u6a2a\u5750\u6807\u4e3ax=-1/2\uff0c
\u628ax=-1/2\u4ee3\u5165y=(\u221a3x-3\uff09/(3/2-\u221a3\uff09\u5f97y=(-\u221a3/2-3\uff09/(3/2-\u221a3\uff09
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=4(2+5\u221a3/4)=8+5\u221a3\u3002\u6240\u4ee5\u70b9C\u7684\u5750\u6807\u4e3aC(-1/2,8+5\u221a3\uff09\u3002
\u4e0b\u9762\u6c42\u70b9A\u7684\u5750\u6807(\u53d6\u5176\u4e2d\u4e00\u79cd\u60c5\u51b5)\uff1a
\u628ay=(\u221a3x-3\uff09/(3/2-\u221a3\uff09\u4ee3\u5165y²=2x\u5f97[(\u221a3x-3\uff09/(3/2-\u221a3\uff09]²=2x,\u5373\uff083x²-6\u221a3x+9)/(21/4-3\u221a3\uff09=2x,3x²-6\u221a3x+9=6(7/4-\u221a3)x,
x²-2\u221a3x+3=2(7/4-\u221a3)x\uff0cx²-7x/2+3=0,x=2\u6216x=3/2(\u70b9B\u7684\u6a2a\u5750\u6807\u4e3a3/2,\u6545\u53d62\u4e3a\u70b9A\u7684\u6a2a\u5750\u6807)\u3002
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=(2\u221a3-3\uff09(3/2+\u221a3\uff09/(-3/4\uff09=4(3-2\u221a3)(3/2+\u221a3\uff09/3=(-6)/3=-2\u3002\u6240\u4ee5\u70b9A\u7684\u5750\u6807\u4e3aA\uff082\uff0c-2\uff09\u3002
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X2+2Y2=18\uff0c\u5373X2/18+Y2/9=1\uff0c
a²=18\uff0cb²=9\uff0c\u5219c²=9\uff0cc=3\uff0c
|F1F2|=2c=6\uff0c\u8bbeP(x0\uff0cy0)\uff0c
S\u25b3F1PF2=1/2*|F1F2|*|y0|=3|y0|\uff0c
\u7531\u5df2\u77e5\u5f97 3|y0|=3\u221a3\uff0c\u5219y0=\u221a3\uff0c
y0²=3\uff0c\u8fd9\u65f6x0²=12.
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\u5219|PF1|\u00b7|PF2|=\uff08a+ex0\uff09\uff08a+ex0\uff09
=a²-e²x0²=18-\uff081/\u221a2\uff09²*12=12.
设AB点：(x1,y1),(x2,y2),设P(x,y)
PA和PB分别与椭圆相切，
所以，PA，PB与OA，OB垂直
斜率相乘为-1
[(y-y1)/(x-x1)]*(y1/x1)=-1
[(y-y2)/(x-x2)]*(y2/x2)=-1
PA与PB垂直
所以，OAPB是矩形，OA与OB垂直
y1y2/(x1x2)=-1
PO=AB,相互平分，
y=y1+y2
x=x1+x2
A,B点代入椭圆方程相加
(x1^2+x2^2)/4+y1^2+Y2^2=2
x^2/4+y^2-2y1y2-x1x2/2=2
y1y2/(x1x2)=-1
x^2/4+y^2+3x1x2/2=2
A,B点代入椭圆方程相减：
(x1^2-x2^2)/4+y1^2-Y2^2=0
(y1-y2)/(x1-x2)=-x/(4y)
[(y1-y2)/(x1-x2)]^2=x^2/(4y)^2
(y^2-4y1y2)/(x^2-4x1x2)=
(y^2+4x1x2)/(x^2-4x1x2)=x^2/(4y)^2
x1x2=(x^4-16y^4)/[4(x^2+16y^2)]
代入：x^2/4+y^2+3x1x2/2=2
得：
P点方程：
5(x^2+4y^2)^2=16(x^2+16y^2)
看不清啊
A,B就是2个轴上端点:答案为4-x^2=y,A（2，0），B（-2，0），设P（x,y）那么有PA垂直PB，所以PA与PB的斜率积为-1，可列：{y/(x+2)}x{y/(x-2)}=-1,可推出：4-x^2=y
你可以问你们老师啊。还有往往你越难做的题越是简单，只不过它的方法在简单的基础上绕了一圈而已，你把数学课本多看看，上面的例题哦。今年高考的时候，我们的数学试卷上就有一个题，本来看似很简单，但没一人做来。回来后翻了一下数学课本，原来是个例题。所以啊，都是对课本不熟悉，那还怎么做题呢。你试试吧，可能对你有帮助。
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