设一抛物线过(0,5),且与抛物线y^2+-4x-2y-3=0共焦点,共轴,试求此抛物线方程。
谢谢帮忙!
參考答案:答案:x + 2 = ( 1 / 8 )( y - 1 )² 或 y² - 8x - 2y - 15 = 0
x - 2 = - ( 1 / 8 )( y - 1 )² 或 y² + 8x - 2y + 17 = 0
详解如下:
y² - 4x - 2y - 3 = 0
4x = y² - 2y - 3
4( x + 1 ) = ( y - 1 )²
x + 1 = ( 1 / 4 )( y - 1 )²
所以抛物线的顶点为 ( - 1, 1 ),其轴为 y = 1
焦点离顶点的距离为 1 / [ 4( 1 / 4 )] = 1
考虑开口向左,所以焦点为 ( 0, 1 )
设所求方程为 x - h = a( y - 1 )²
则 0 - h = a( 5 - 1 )² => h = - 16a
及 | h - 0 | = 1 / ( 4a ) => h = ± 1 / ( 4a )
- 16a = ± 1 / ( 4a )
64a² = ± 1
a² = 1 / 64 或 a² = - 1 / 64 ( 舍去 )
a = ± 1 / 8
若 a = 1 / 8,h = - 2;若 a = - 1 / 8,h = 2
所以所求方程为 x + 2 = ( 1 / 8 )( y - 1 )² 或 x - 2 = - ( 1 / 8 )( y - 1 )²