5.2 Parabola 抛物线 - 独中课程 | 线上补习

高三数学 | 高级数学

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5.2 Parabola 抛物线

1. Find the vertex, focus, directrix of the following parabola and sketch the graph.
试求以下方程式之顶点, 焦点, 准线並描其图形.
(a) $y^2 = -8x$
(b) $x^2 = 16y$
(c) $y^2 - 8x - 2y - 7 = 0$
(d) $x^2 + 4x + 16y - 44 = 0$

Answer 1(a) ：

Answer 1(b) ：

Answer 1(c) ：

Answer 1(d) ：

2. Find the equation of the parabola with vertex at (5, -2) and focus at (5, -4)

求顶点为(5, -2)及焦点为(5, -4)的抛物线方程式

Answer:

$x^2 - 10x + 8y + 41 = 0$

3. A parabola has its axis parallel to the y-axis, one end of its latus rectum is at (9, 6) and the vertex is at (5, 4). Find

一抛物线其对称轴平行于y轴, 通径的一端在(9, 6)且顶点在(5, 4). 求

(a) the length of the latus rectum
其通径长度

(b) the equation of the parabola
该抛物线的方程式

Answer :

(a) 8

(b) x $^2$ – 10x – 8y + 57 = 0

4. Find the equation of a parabola where the vertex of the parabola is (-2, 3), the focus is on the line 4x – 3y + 5 = 0 and the symmetric axis is parallel to

求一拋物线方程式其中该拋物线的顶点坐标是(－2, 3)，焦点在直线4x – 3y + 5 = 0上且对称轴平行于

(a) x – axis,
x-轴

(b) y – axis
y-坐标轴，

Answer：

(a) $(y - 3)^2 = 12(x + 2)$

(b) $(x + 2)^2 = -16(y - 3)$

5. Find the equation of a parabola with vertex on the line y = 2x, symmetric axis parallel to the x-axis and passing through (1½, 1 ) and (3, 4)

求一抛物线的方程式其顶点在直线y = 2x上, 对称轴平行于x轴且经过点 (1½, 1) 及 (3, 4).

Answer：

$(y - 2)^2 = 2(x - 1)$ , $(y - 7)^2 = -8(x - \frac{7}{2} )$

6. Find the equation of a parabola through (-1, -3) and (2, 1.5), whose axis is parallel to the y-axis and latus rectum equal to 6.

求经过 (-1, -3) 和 (2, 1.5), 其对称轴平行于y轴及通径长等于6的抛物线方程式.

Answer：

$(x - 5)^2 = -6(y - 3)$ , $(x + 4)^2 = 6(y + \frac{9}{2})$

7. A parabola has directrix x = 6, axis y = -5 and latus rectum equal to 9. Find its equation.

一抛物线其准线为x = 6, 对称为y = -5 和通线长等于9, 求其方程式.

Answer：

$(y + 5)^2 = -9(x - 3\frac{3}{4})$ , $(y + 5)^2 = 9(x - 8\frac{1}{4})$