直线y=x+m与椭圆x^2/16+y^2/9=1相交 求m的取值范围X平方/16+Y平方/9=1 整理的:9X^2+16Y^2-144=0带Y=X+M入椭圆方程,得:9X^2+16(X+M)^2-144=0整理的:25X^2+32MX+16M^2-144=0因为有两个交点,所以Δ>=0即,Δ
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![直线y=x+m与椭圆x^2/16+y^2/9=1相交 求m的取值范围X平方/16+Y平方/9=1 整理的:9X^2+16Y^2-144=0带Y=X+M入椭圆方程,得:9X^2+16(X+M)^2-144=0整理的:25X^2+32MX+16M^2-144=0因为有两个交点,所以Δ>=0即,Δ](/uploads/image/z/6931291-67-1.jpg?t=%E7%9B%B4%E7%BA%BFy%3Dx%2Bm%E4%B8%8E%E6%A4%AD%E5%9C%86x%5E2%2F16%2By%5E2%2F9%3D1%E7%9B%B8%E4%BA%A4+%E6%B1%82m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4X%E5%B9%B3%E6%96%B9%EF%BC%8F16%EF%BC%8BY%E5%B9%B3%E6%96%B9%EF%BC%8F9%EF%BC%9D1+%E6%95%B4%E7%90%86%E7%9A%84%EF%BC%9A9X%5E2%EF%BC%8B16Y%5E2-144%EF%BC%9D0%E5%B8%A6Y%3DX%2BM%E5%85%A5%E6%A4%AD%E5%9C%86%E6%96%B9%E7%A8%8B%2C%E5%BE%97%EF%BC%9A9X%5E2%EF%BC%8B16%28X%2BM%29%5E2-144%EF%BC%9D0%E6%95%B4%E7%90%86%E7%9A%84%EF%BC%9A25X%5E2%2B32MX%2B16M%5E2-144%3D0%E5%9B%A0%E4%B8%BA%E6%9C%89%E4%B8%A4%E4%B8%AA%E4%BA%A4%E7%82%B9%2C%E6%89%80%E4%BB%A5%CE%94%3E%3D0%E5%8D%B3%2C%CE%94)
直线y=x+m与椭圆x^2/16+y^2/9=1相交 求m的取值范围X平方/16+Y平方/9=1 整理的:9X^2+16Y^2-144=0带Y=X+M入椭圆方程,得:9X^2+16(X+M)^2-144=0整理的:25X^2+32MX+16M^2-144=0因为有两个交点,所以Δ>=0即,Δ
直线y=x+m与椭圆x^2/16+y^2/9=1相交 求m的取值范围
X平方/16+Y平方/9=1 整理的:9X^2+16Y^2-144=0
带Y=X+M入椭圆方程,得:9X^2+16(X+M)^2-144=0
整理的:25X^2+32MX+16M^2-144=0
因为有两个交点,所以Δ>=0
即,Δ=(32M)^2-4*25*(16M^2-144)>=0
整理得:M^2-25
直线y=x+m与椭圆x^2/16+y^2/9=1相交 求m的取值范围X平方/16+Y平方/9=1 整理的:9X^2+16Y^2-144=0带Y=X+M入椭圆方程,得:9X^2+16(X+M)^2-144=0整理的:25X^2+32MX+16M^2-144=0因为有两个交点,所以Δ>=0即,Δ
Δ=(32M)^2-4*25*(16M^2-144)=32²M²-25*64M+25*64*9>=0,两边约去64得16M²-25M²+25*9>=0
即M²-25
Δ=(32M)^2-4*25*(16M^2-144)>=0
1024m^2-1600(m^2-9)>=0
16m^2-25(m^2-9)>=0
-9m^2+225>=0
25-m^2>=0
-5<=M<=5
另外本题如果相交的话 Δ>0即可
Δ=0含有相切这种情况
(32m)^2-4*25(16m^2-144)>=0
1024m^2-1600m^2+14400>=0
-576m^2+14400>=0
576(-m^2+25)>=0
-m^2+25>=0
m^2-25<=0
Δ=(32M)^2-4*25*(16M^2-144)=322M2-25*64M+25*64*9>=0,两边约去64得16M2-25M2+25*9>=0
即M2-25<=0,故-5<=M<=5