在以下每个练习1至6中，找到焦点的坐标，抛物线的轴，直角的方程式和直肠的长度。
问题1：y 2 = 12x
解决方案：
Given equation: y2 = 12x
Since the coefficient of x is positive.
Therefore, the parabola will open towards the right.
While comparing this equation with y2 = 4ax, we get,
4a = 12
a = 3
Therefore, the co-ordinates of the focus = (a, 0) = (3, 0)
Since, the given equation involves y2,
The axis of the parabola is the x-axis.
Thus, the equation of directrix, x = -a, therefore
x = -3
The length of latus rectum = 4a = 4 × 3 = 12
问题2：x 2 = 6y
解决方案：
Given equation : x2 = 6y
Since the coefficient of y is positive.
Therefore, the parabola will open upwards.
While comparing this equation with x2 = 4ay, we get,
4a = 6
a = 6/4
= 3/2
Therefore, the co-ordinates of the focus = (0, a) = (0, 3/2)
Since, the given equation involves x2,
The axis of the parabola is the y-axis.
Thus, the equation of directrix, y =-a, therefore,
y = -3/2
The length of latus rectum = 4a = 4(3/2) = 6
问题3：y 2 = – 8x
解决方案：
Given equation: y2 = -8x
Since the coefficient of x is negative.
Therefore, the parabola will open towards the left.
While comparing this equation with y2 = -4ax, we get,
-4a = -8
a = -8/-4 = 2
Therefore, the co-ordinates of the focus = (-a,0) = (-2, 0)
Since, the given equation involves y2,
The axis of the parabola is the x-axis.
Thus, the equation of directrix, x = a, therefore,
x = 2
The length of latus rectum = 4a = 4 (2) = 8
问题4：x 2 = – 16y
解决方案：
Given equation: x2 = -16y
Since the coefficient of y is negative.
Therefore, the parabola will open downwards.
While comparing this equation with x2 = -4ay, we get,
-4a = -16
a = -16/-4
= 4
Therefore, the co-ordinates of the focus = (0,-a) = (0,-4)
Since, the given equation involves x2,
The axis of the parabola is the y-axis.
Thus, the equation of directrix, y =a, then,
y = 4
The length of latus rectum = 4a = 4(4) = 16
问题5：y 2 = 10x
解决方案：
Given equation: y2 = 10x
Since the coefficient of x is positive.
Therefore, the parabola will open towards the right.
While comparing this equation with y2 = 4ax, we get,
4a = 10
a = 10/4 = 5/2
Therefore, co-ordinates of the focus = (a, 0) = (5/2, 0)
Since, the given equation involves y2,
The axis of the parabola is the x-axis.
Thus, the equation of directrix, x = -a, then,
x = – 5/2
The length of latus rectum = 4a = 4(5/2) = 10
问题6：x 2 = – 9y
解决方案：
Given equation: x2 = -9y
Since the coefficient of y is negative.
Therefore, the parabola will open downwards.
While comparing this equation with x2 = -4ay, we get,
-4a = -9
a = -9/-4 = 9/4
Therefore, co-ordinates of the focus = (0,-a) = (0, -9/4)
Since, the given equation involves x2,
The axis of the parabola is the y-axis.
Thus, the equation of directrix, y = a, then,
y = 9/4
The length of latus rectum = 4a = 4(9/4) = 9
在练习7至12中，找到满足给定条件的抛物线方程：
问题7：重点(6,0); Directrix x = – 6
解决方案：
Given: Focus (6,0) and directrix x = -6
Since the focus lies on the x–axis
The x-axis is the axis of the parabola.
As the directrix, x = -6 thus it is to the left of the y- axis,
The equation of the parabola is y2 = 4ax
Here, a = 6
Therefore, the equation of the parabola is y2 = 24x.
问题8：重点(0，–3)；方向y = 3
解决方案：
Given: Focus (0, -3) and directrix y = 3
Since the focus lies on the y–axis,
The y-axis is the axis of the parabola.
As the given directrix, y = 3 thus it is above the x- axis,
The equation of the parabola is x2 = -4ay
Here, a = 3
Therefore, the equation of the parabola is x2 = -12y.
问题9：顶点(0，0);焦点(3，0)
解决方案：
Given: Vertex (0, 0) and focus (3, 0)
Since the vertex of the parabola is (0, 0)
The focus lies on the positive x-axis.
The parabola is of the form y2 = 4ax.
Since, the focus is (3, 0), a = 3
Therefore, the equation of the parabola is y2 = 4 × 3 × x,
y2 = 12x
问题10：顶点(0，0);焦点(–2，0)
解决方案：
Given: Vertex (0, 0) and focus (-2, 0)
Since the vertex of the parabola is (0, 0)
The focus lies on the positive x-axis.
The parabola is of the form y2=-4ax.
Since, the focus is (-2, 0), a = 2
Therefore, the equation of the parabola is y2 = -4 × 2 × x,
y2 = -8x
问题11：顶点(0，0)穿过(2，3)，并且轴沿x轴。
解决方案：
Given: Vertex is (0, 0) and the axis is along the x-axis
Also given that the parabola passes through the point (2, 3), which lies in the first quadrant.
Since equation of the parabola is y2 = 4ax while the point (2, 3) must satisfy the equation y2 = 4ax.
Therefore,
32 = 4a(2)
32 = 8a
9 = 8a
a = 9/8
Thus, the equation of the parabola is
y2 = 4 (9/8)x
= 9x/2
2y2 = 9x
Therefore, the equation of the parabola is 2y2 = 9x
问题12：顶点(0，0)，穿过(5，2)并相对于y轴对称。
解决方案：
Given: Vertex is (0, 0) and the symmetric is with respect to the y-axis.
Also given that the parabola passes through the point (5, 2), which lies in the first quadrant.
Since, the equation of the parabola is x2 = 4ay while the point (5, 2) must satisfy the equation x2 = 4ay.
Therefore,
52 = 4a(2)
25 = 8a
a = 25/8
Thus, the equation of the parabola is
x2 = 4 (25/8)y
x2 = 25y/2
2x2 = 25y
Therefore, the equation of the parabola is 2x2 = 25y