Question

已知椭圆中心在原点,焦点在横轴上,焦距为4,且和直线3x+2y-16=0相切,求椭圆方程.

Answer 1

椭圆的方程为+=1.

解法一:设椭圆方程为=1(a>b>0),
切点为P(x₀,y₀),则切线为="1.                                       " ①
又切线为3x+2y-16="0,                                                    " ②
故直线①②重合.
∴,即x₀=,y₀=b².
代入②,得9a²+28b²-256="0.                                                   " ③
又焦距为4,∴c=2.
∴a²-b²="12.                                                                 " ④
联立方程③④,解得
故所求椭圆的方程为+=1.
解法二:c=2,c²=12.
设椭圆=1与直线方程联立,Δ=0得b²=4.
故所求椭圆的方程为+=1.