同构处理,计算量大,弦长问题
已知\(A(2,2),B,C\)是抛物线\(E:x^2=2py\)上的三点,且\(AB\)与直线\(AC\)的斜率和\(0\)
(1)求直线\(BC\)的斜率
(2)若直线\(AB,AC\)均与圆\(M:x^2+(y-2)^2=r^2(0<r<\sqrt{3})\)相切,且直线\(BC\)被圆\(M\)所截得的线段长\(\dfrac{2\sqrt{30}}{5}\),求\(r\)的值
解
(1)略,\(p=1,k_{BC}=-2\)
(2)设\(B\left(x_1,\dfrac{x_1^2}{2}\right),C\left(x_2,\dfrac{x_2^2}{2}\right),l_{BC}:y=-2x+m\),与抛物线联立\(\begin{cases} y=-2x+m\\ x^2=2y \end{cases}\)
得\(x^2+4x-m=0\),即\(\Delta=16+8m>0,\begin{cases} x_1+x_2=-4\\ x_1x_2=-2m \end{cases}\)
圆心到\(l_{BC}\)的距离\(d=\dfrac{|2-m|}{\sqrt{5}}\)
根据弦长关系:\(r^2-\dfrac{(2-m)^2}{5}=\left(\dfrac{\sqrt{30}}{5}\right)^2\)①
设\(AB:y-2=\dfrac{2-\dfrac{x_1^2}{2}}{2-x_1}(x-2)\)即\(y-\dfrac{2+x_1}{2}(x-2)-2=0\)
因\(AB\)与圆相切\(\dfrac{|2+x_1|}{\sqrt{1+\dfrac{(2+x_1)^2}{4}}}=r\)即\(\dfrac{|4+2x_1|}{\sqrt{4+(x_1+2)^2}}=r\)
整理有:\(4x_1^2+16x_1+16=4r^2+r^2+x_1^2+4r^2x_1+4r^2\)
再整理:\(x_1^2(4-r^2)(15-4r^2)x_1+16-8r^2=0\)
再整理:\(x_1^2+4x_1+\dfrac{16-8r^2}{4-r^2}=0\)
同理\(C\)点一样满足:\(x_2^2+4x_2+\dfrac{16-8r^2}{4-r^2}=0\)
则\(x_1,x_2\)是一元二次方程
\(x^2+4x+\dfrac{16-8r^2}{4-r^2}=0\)的两个根
则\(x_1x_2=\dfrac{16-8r^2}{4-r^2}\)
而\(x_1x_2=-2m\)
则有
\(\dfrac{16-8r^2}{4-r^2}=2m\)②
联立①,②有\(\begin{cases} \dfrac{16-8r^2}{4-r^2}=2m\\ r^2-\dfrac{(2-m)^2}{5}=\left(\dfrac{\sqrt{30}}{5}\right)^2 \end{cases}\),即\(\begin{cases} 5r^2-(2-m)^2=6\\ r^2=\dfrac{8+4m}{4+m} \end{cases}\)
即$$40+20m-(2-m)^2(4+m)=6m-16$$
整理有\(m^3=26m\)
得\(m=0,\pm\sqrt{26}\),因\(0<r<\sqrt{3}\)
从而\(m=0,r=\sqrt{2}\).