Question

已知直线y=x-1与椭圆4分之x²+2分之y²=1交于AB两点，求AB的长

Answer 1

已知直线y=x-1与椭圆x²/4+y²/2=1交于AB两点，求AB的长
解：将y=x-1代入椭圆方程得x²+2(x-1)²-4=3x²-4x-2=0
设A(x₁，y₁)；B(x₂，y₂)；则x₁+x₂=4/3；x₁x₂=-2/3；
故∣AB∣=√{2[(x₁+x₂)²-4x₁x₂]}=√[2[(16/9)+(8/3)]=√(80/9)=(4/3)√5.

Answer 2

设a，b两点坐标为(x1，y1)，(x2，y2)将y
=
x
-1代入x²/5
+
y²/4
=
1;
得x²/5
+
(x-1)²/4
=
1;
即9x²
-
10x
-15
=
0；
x1+x2=10/9;
x1x2
=
-
5/3;
|ab|
=
√[(x1-x2)²+(y1-y2)²]
=
√2√[(x1+x2)²
-
4x1x2]
=
16√5/9;