一道解析几何题目,求多解已知椭圆C:x²/4+y²/3=1,圆O:x²+y²=7

解1 设P(p,q),则q^2=7-p^2.①
直线l:y=kx+q-kp②与椭圆x^2/4+y^2/3=1③相切,
把②代入③*12,得
3x^2+4[k^2x^2+2k(q-kp)x+(q-kp)^2]=12,
整理得(3+4k^2)x^2+8k(q-kp)x+4(q-kp)^2-12=0,
△/16=4k^2*(q-kp)^2-(3+4k^2)[(q-kp)^2-3]
=-[3(q-kp)^2-9-12k^2]=0,
∴q^2-2kpq+k^2p^2-3-4k^2=0,
整理得(p^2-4)k^2-2kpq+4-p^2=0,(由①)
设切线l1,l2的斜率分别是k1,k2,则k1k2=-1,
∴l1⊥l2.
把l:y=kx+q-kp④
代入圆：x^2+y^2=7得x^2+(kx+q-kp)^2=7,
整理得(1+k^2)x^2+2k(q-kp)x+(q-kp)^2-7=0,
解得x1=xP=p,x2=xM=[(q-kp)^2-7]/[p(1+k^2)],
X1-x2=[(1+k^2)p^2-(q-kp)^2+7]/[p(1+k^2)]
=(7+p^2-q^2+2kpq)/[p(1+k^2)]=(2p+2kq)/(1+k^2),
∴PM^2=|x1-x2|^2*(1+k^2)=4(p+kq)^2/(1+k^2),
以-1/k代k,得PN^2=4(p-q/k)^2/[1+(-1/k)^2]=4(kp-q)^2/(k^2+1),
∴MN^2=PM^2+PN^2=4/(1+k^2)*[(p+kq)^2+(kp-q)^2]
=4/(1+k^2)*(1+k^2)(p^2+q^2)=28,（由①）
∴|MN|=2√7.
解2 设P(√7cost,√7sint),
切线l：y-√7sint=k(x-√7cost),即y=kx+√7(sint-kcost),①
把①代入x^2/4+y^2/3=1,②得
3x^2+4[k^2x^2+2√7k(sint-kcost)x+7(sint-kcost)^2]=12,
整理得(3+4k^2)x^2+8√7k(sint-kcost)x+28(sint-kcost)^2-12=0,
△/16=28k^2*(sint-kcost)^2-(3+4k^2)[7(sint-kcost)^2-3]
=-[21(sint-kcost)^2-9-12k^2]=0,
∴7[(sint)^2-2ksintcost+k^2*(cost)^2]-3-4k^2=0,
整理得[7(cost)^2-4]k^2-14ksintcost+4-7(cost)^2=0,
设切线l1,l2的斜率分别是k1,k2,则k1k2=-1,
∴l1⊥l2.余者仿上.