若椭圆x²/a²+y²/b²=1(a>b>1)内有圆x²+y²=1,该圆的切线与椭圆交于A、B两点,且满足向量OA·向量OB=0,则9a²+16b²的最小值为...
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![若椭圆x²/a²+y²/b²=1(a>b>1)内有圆x²+y²=1,该圆的切线与椭圆交于A、B两点,且满足向量OA·向量OB=0,则9a²+16b²的最小值为...](/uploads/image/z/12052811-11-1.jpg?t=%E8%8B%A5%E6%A4%AD%E5%9C%86x%26%23178%3B%2Fa%26%23178%3B%2By%26%23178%3B%2Fb%26%23178%3B%3D1%EF%BC%88a%EF%BC%9Eb%EF%BC%9E1%EF%BC%89%E5%86%85%E6%9C%89%E5%9C%86x%26%23178%3B%2By%26%23178%3B%3D1%2C%E8%AF%A5%E5%9C%86%E7%9A%84%E5%88%87%E7%BA%BF%E4%B8%8E%E6%A4%AD%E5%9C%86%E4%BA%A4%E4%BA%8EA%E3%80%81B%E4%B8%A4%E7%82%B9%2C%E4%B8%94%E6%BB%A1%E8%B6%B3%E5%90%91%E9%87%8FOA%C2%B7%E5%90%91%E9%87%8FOB%3D0%2C%E5%88%999a%26%23178%3B%2B16b%26%23178%3B%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA...)
若椭圆x²/a²+y²/b²=1(a>b>1)内有圆x²+y²=1,该圆的切线与椭圆交于A、B两点,且满足向量OA·向量OB=0,则9a²+16b²的最小值为...
若椭圆x²/a²+y²/b²=1(a>b>1)内有圆x²+y²=1,该圆的切线
与椭圆交于A、B两点,且满足向量OA·向量OB=0,则9a²+16b²的最小值为...
若椭圆x²/a²+y²/b²=1(a>b>1)内有圆x²+y²=1,该圆的切线与椭圆交于A、B两点,且满足向量OA·向量OB=0,则9a²+16b²的最小值为...
如附件,总之一个思想就是利用直线与圆或者椭圆的方程的交点(方程的根)来解题,并且利用向量垂直时的乘积为0.
圆的切线:xcosu+ysinu=1,
即y=(1-xcosu)/sinu,①
代入椭圆方程:b^2x^2+a^2y^2=a^2b^2得
b^2x^2+a^2*(1-xcosu)^2/(sinu)^2=a^2b^2,
∴b^2x^2(sinu)^2+a^2*[1-2xcosu+x^2(cosu)^2]=a^2b^2(sinu)^2,
整理得[b^2(sinu...
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圆的切线:xcosu+ysinu=1,
即y=(1-xcosu)/sinu,①
代入椭圆方程:b^2x^2+a^2y^2=a^2b^2得
b^2x^2+a^2*(1-xcosu)^2/(sinu)^2=a^2b^2,
∴b^2x^2(sinu)^2+a^2*[1-2xcosu+x^2(cosu)^2]=a^2b^2(sinu)^2,
整理得[b^2(sinu)^2+a^2(cosu)^2]x^2-2a^2xcosu+a^2-a^2b^2(sinu)^2=0,
设A(x1,y1),B(x2,y2),则x1+x2=2a^2*cosu/[b^2(sinu)^2+a^2(cosu)^2],
x1x2=[a^2-a^2b^2(sinu)^2]/[b^2(sinu)^2+a^2(cosu)^2],
由①,y1y2=(1-x1cosu)(1-x2cosu)/(sinu)^2
=[1-(x1+x2)cosu+x1x2(cosu)^2]/(sinu)^2,
由0=向量OA*OB=x1x2+y1y2得
x1x2-(x1+x2)cosu+1=0,
∴a^2-a^2b^2(sinu)^2-2a^2*(cosu)^2+b^2(sinu)^2+a^2(cosu)^2=0,
∴a^2+b^2=a^2b^2,b^2=a^2/(a^2-1),
∴9a^2+16b^2=9a^2+16+16/(a^2-1)
=9(a^2-1)+16/(a^2-1)+25
>=24+25=49,当a^2-1=4/3,a^2=7/3,b^2=7/4时取等号,
∴所求最小值=49.
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设切线的斜率为k,方程为y=kx+b,
——》原点到直线的距离h=b/√(k^2+1)=1,
——》b=√(k^2+1),即切线方程为:y=kx+√(k^2+1),
代入椭圆方程:x^2/a^2+[kx+√(k^2+1)]^2/b^2=1,
整理得:(k^2+b^2/a^2)x^2+2k√(k^2+1)x+k^2+1-b^2=0,
——》xa*xb=(k^2...
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设切线的斜率为k,方程为y=kx+b,
——》原点到直线的距离h=b/√(k^2+1)=1,
——》b=√(k^2+1),即切线方程为:y=kx+√(k^2+1),
代入椭圆方程:x^2/a^2+[kx+√(k^2+1)]^2/b^2=1,
整理得:(k^2+b^2/a^2)x^2+2k√(k^2+1)x+k^2+1-b^2=0,
——》xa*xb=(k^2+1-b^2)/(k^2+b^2/a^2),
xa+xb=-2k√(k^2+1)/(k^2+b^2/a^2),
ya*yb=[kxa+√(k^2+1)][kxb+√(k^2+1)]=k^2xaxb+k√(k^2+1)(xa+xb)+k^2+1,
向量OA·向量OB=0,——》xa*xb+ya*yb=0,
——》(k^2+1)xaxb+k√(k^2+1)(xa+xb)+k^2+1=0,
——》(k^2+1)(k^2+1-b^2)/(k^2+b^2/a^2)-2k^2(k^2+1)/(k^2+b^2/a^2)+(k^2+1)=0
整理得:1/a^2+1/b^2=1,
令1/a^2=s,1/b^2=t,则:
s+t=1,——》ds+dt=0,
m=9a^2+16b^2=9/s+16/t,——》dm=-9ds/s^2-16dt/t^2=(9/s^2-16/t^2)dt,
m取最小值,则:dm=0,——》9/s^2-16/t^2=9/s^2-16/(1-s)^2=0,
整理得:7s^2+18s-9=(s+3)(7s-3)=0,
——》s=3/7,——》t=4/7,
——》m最小=9/s+16/t=49,
即9a^2+16b^2的最小值为49。
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