解析:设直线L1:y=x+b与椭圆x2/16+y2/9=1相切于点D,
代入得25x^2+32bx+16b^2-144=0,
△=(32b)^2-4*25*(16b^2-144)=0
解得b=-5,b=5(舍去),有y-x+5=0
得x=16/5,y=-9/5,
即D(16/5,-9/5)
d=│-9/5-16/5-5│/√2=5√2,
令x=0,y=5,y=0.x=-5,
即A(-5,0),B(0,5)
∴AB=5√2,
S△ABD=1/2*AB*d=1/2*5√2*5√2=25
解析:设直线L1:y=x+b与椭圆x2/16+y2/9=1相切于点D,
代入得25x^2+32bx+16b^2-144=0,
△=(32b)^2-4*25*(16b^2-144)=0
解得b=-5,b=5(舍去),有y-x+5=0
得x=16/5,y=-9/5,
即D(16/5,-9/5)
d=│-9/5-16/5-5│/√2=5√2,
令x=0,y=5,y=0.x=-5,
即A(-5,0),B(0,5)
∴AB=5√2,
S△ABD=1/2*AB*d=1/2*5√2*5√2=25