如何在给定焦点和短轴的情况下找到椭圆的方程?
圆锥截面,通常称为圆锥曲线,是在平面与圆锥相交时形成的。这些部分相交的角度决定了它们的形状。因此,圆锥截面分为四种类型:圆、椭圆、抛物线和双曲线。这些类型中的每一种都有自己的一组方程和数学属性。下面讨论椭圆。
椭圆
椭圆是当平面以小于直角但大于在圆锥顶点处形成的角度 (α) 的角度 (β) 与圆锥相交时生成的圆锥截面。换句话说,当平面以角度 β 切割圆锥体时生成椭圆,使得 α<β<90 o 。
如上图所示,锥体和平面以小于直角但大于 α 的角度 β 相交以产生椭圆。
椭圆方程
- 以 (h, k) 为中心且长轴平行于 x 轴的椭圆的标准方程由下式给出:
,
where the coordinates of the vertex are (h±a, 0), coordinates of co-vertex are (h, k±b) and the coordinates of foci are (h±c, k), where c2 = a2 – b2.
- 以 (h, k) 为中心且长轴平行于 y 轴的椭圆的标准方程由下式给出:
,
where the coordinates of the vertex are (h, k±a), coordinates of co-vertex are (h±b, k) and the coordinates of foci are (h, k±c), where c2 = a2 – b2.
如何在给定焦点和短轴的情况下找到椭圆的方程?
解决方案:
To find the equation of an ellipse, we need the values a and b. Now, we are given the foci (c) and the minor axis (b). To calculate a, use the formula c2 = a2 – b2. Substitute the values of a and b in the standard form to get the required equation.
Let us understand this method in more detail through an example.
Example: Say, an ellipse centered at origin with foci (±4, 0) and minor axis (0, ±3).
Given b = 3 and c = 4.
Put these in the formula c2 = a2 – b2 to find a.
a2 = 32 + 42
a2 = 25
a = 5
As the ellipse lies on x-axis, the equation is of the form .
So, the equation is, .
类似问题
问题 1. 找出以原点为中心、焦点 (±7, 0) 和短轴 (0, ±5) 的椭圆的方程。
解决方案:
Given b = 5 and c = 7.
Put these in the formula c2 = a2 – b2 to find a.
a2 = 52 + 72
a2 = 74
As the ellipse lies on x-axis, the equation is of the form
So, the equation is,
问题 2. 找出以原点为中心、焦点 (0, ±5) 和短轴 (12, 0) 的椭圆的方程。
解决方案:
Given b = 12 and c = 5.
Put these in the formula c2 = a2 – b2 to find a.
a2 = 52 + 122
a2 = 169
a = 13
As the ellipse lies on y-axis, the equation is of the form
So, the equation is,
问题 3. 找出以 (3, 2) 为中心且 c = 6 且 b = 8 的椭圆的方程。
解决方案:
Given b = 8, c = 6, h = 3 and k = 2.
Put these in the formula c2 = a2 – b2 to find a.
a2 = 82 + 62
a2 = 100
a = 10
As the ellipse lies on x-axis, the equation is of the form
So, the equation is,
问题 4. 用焦点 (0, ±5) 和短轴 (12, 0) 找到椭圆的长轴坐标。
解决方案:
We have, c = 5 and b = 12.
Put these in c2 = a2 – b2 to find a.
a2 = 122 + 52
a2 = 169
a = 13
The coordinates of major axis are (0, ±13).
问题 5. 用焦点 (±24, 0) 和短轴 (0, 10) 找到椭圆长轴的坐标。
解决方案:
We have, c = 24 and b = 10.
Put these in c2 = a2 – b2 to find a.
a2 = 102 + 242
a2 = 676
a = 26
The coordinates of major axis are (0, ±26).