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


(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\(\begin{cases} y=-2x+m\\ x^2=2y \end{cases}\)

\(x^2+4x-m=0\),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:\(r^2-\dfrac{(2-m)^2}{5}=\left(\dfrac{\sqrt{30}}{5}\right)^2\)

\(AB:y-2=\dfrac{2-\dfrac{x_1^2}{2}}{2-x_1}(x-2)\)\(y-\dfrac{2+x_1}{2}(x-2)-2=0\)

\(AB\)与圆相切\(\dfrac{|2+x_1|}{\sqrt{1+\dfrac{(2+x_1)^2}{4}}}=r\)\(\dfrac{|4+2x_1|}{\sqrt{4+(x_1+2)^2}}=r\)

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:\(x_2^2+4x_2+\dfrac{16-8r^2}{4-r^2}=0\)

\(x_1,x_2\)是一元二次方程
\(x^2+4x+\dfrac{16-8r^2}{4-r^2}=0\)的两个根

\(x_1x_2=\dfrac{16-8r^2}{4-r^2}\)

\(x_1x_2=-2m\)

则有
\(\dfrac{16-8r^2}{4-r^2}=2m\)

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\(m^3=26m\)

\(m=0,\pm\sqrt{26}\),因\(0<r<\sqrt{3}\)

从而\(m=0,r=\sqrt{2}\).

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