Question

雙曲線C的漸近線方程為x±2y=0,點(diǎn)A(5,0)到雙曲線C上動(dòng)點(diǎn)P的距離的最小值為6.

(1)求雙曲線方程;

(2)若過(guò)B(1,0)點(diǎn)的直線l交雙曲線C上支一點(diǎn)M,下支一點(diǎn)N,且4MB=5BN,求直線l的方程.

Answer 1

解:(1)若雙曲線焦點(diǎn)在x軸上，?

∵漸近線方程為y=±x,?

∴雙曲線方程設(shè)為=1(b＞0).                                     ?

設(shè)動(dòng)點(diǎn)P的坐標(biāo)為(x,y),?

則|AP|=.?

∵x∈(-∞，-2b］∪［2b,+∞),?

∴①若x=4≤2b,即b≥2,則當(dāng)x=2b時(shí),|AP|_Min=|2b-5|=.?

解得b=(b=＜2應(yīng)舍去)，?

此時(shí)雙曲線方程為-=1.                                ?

②若x=4＞2b,即b＜2，則當(dāng)x=4時(shí),|AP|_Min=

∴b²=-1，無(wú)解.                                                         ?

若雙曲線焦點(diǎn)在y軸上，雙曲線方程可設(shè)為=1(b＞0),?

∴|PA|=.?

∵x∈R,∴x=4時(shí),|PA|_Min=.∴b=1.?

此時(shí)雙曲線方程為y²-=1.?

綜上所述,雙曲線方程為-=1或y²-=1.                ?

(2)由(1)知,雙曲線方程為y²-=1，設(shè)直線l方程為x=ky+1.?

由得(4-k²)y²-2ky-5=0,依題意

-＜0.                                                                           

∴-2＜k＜2.設(shè)M(x₁,y₁),y₁＞0,N(x₂,y₂),y₂＜0,?

由韋達(dá)定理得y₁+y₂=;                                               ①?

y₁·y₂=-.                                                               ②?

∵4=5,∴-4y₁=5y₂.                                                ③

由③得y₁=-y₂,代入①②得?

-y₂=＞0,                                                              ④?

-y₂²=-,                                                         ⑤?

即ln＜=1+(n≥2).?

∴l(xiāng)n+ln+ln+…+ln＜(1+)+(1+)+…+(1+)=n+++…+.?

綜上所證,++…+＜lnn＜n+++…+(n∈N^*且n≥2)成立.       ?

由④⑤消去y₂,解得k=(k=-＜0,舍).?

∴直線l的方程為x=y+1.