拋物線y=x2+4x上一點P處的切線的傾斜角為45°,切線與x,y軸的交點分別是A,B,則△AOB的面積為 .
【答案】
分析:由題意和導(dǎo)數(shù)的幾何意義求出點P的坐標,再求出切線方程,然后求出A、B兩點的坐標,進而可求長度及直線AB的方程,再求原點到AB得距離即為三角形邊AB上的高,再代入三角形的面積公式求解.
解答:解:設(shè)點P的坐標為(x,y),
由題意,y'=2x+4且過P點的切線的斜率k=tan45°=1,
∴由導(dǎo)數(shù)的幾何意義得,1=2x+4,x=-

;代入y=x
2+4x解得,y=-

,
∴P的坐標為(-

,-

),
∴過P點的切線的方程為y+

=x+

,即x-y-

=0,
令y=0,x=

,令x=0,y=-

;∴A(

,0),B(0,-

)
∴|AB|=

=

,直線AB的方程為x-y-

=0;
∴點O(0,0)到直線AB的方程得距d=

=

,
∴△AOB的面積S=

×|AB|×d=

.故答案為:

.
點評:本題考查了根據(jù)導(dǎo)數(shù)的幾何意義如何求切點和切線方程,還有直線方程及三角形的面積求法,是一道好題.