如何找到具有焦点和点的椭圆方程？

当平面穿过圆锥时，会创建圆锥截面，也称为圆锥曲线。这些部分的几何形状由它们交叉的角度决定。因此，圆锥曲线分为四类：圆、椭圆、抛物线和双曲线。这些形式中的每一种都有自己的一组数学特征和方程。下面讨论椭圆。

椭圆

作为圆锥截面，椭圆是当平面以小于直角但大于在圆锥顶点处形成的角（α）的角度（β）与圆锥相交时形成的形状。换言之，当平面以角度β切割圆锥时形成椭圆，使得α<β<90 ^° 。

如上图所示，圆锥与平面以小于直角但大于α的角β相交，因相交而形成椭圆。

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：

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$, 

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.

水平椭圆

• 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：

$\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ , 

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.

垂直椭圆

如何找到具有焦点和点的椭圆方程？

解决方案：

To find the equation of an ellipse, we need the values a and b. Now, it is known that the sum of the distances of a point lying on an ellipse from its foci is equal to the length of its major axis, 2a. The value of a can be calculated by this property. To calculate b, use the formula c² = a² – b². Substitute the obtained 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 passing through the origin with foci (±4, 0) and point (–4, 1.8).

Using the formula, we have

$2a=1.8+\sqrt{(4-(-4))^2+(0-1.8)^2}$

2a = 10

a = 5

Put a = 5 in c² = a² – b² to find b.

b² = 25 – 16

b² = 9 

As the ellipse lies on x-axis, the equation is of the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$.

So, the equation is, $\frac{x^2}{25}+\frac{y^2}{9}=1$.

类似问题

问题 1. 求一个椭圆的方程，它通过焦点 (±7, 0) 和点 (6, 2) 的原点。

解决方案：

Using the formula, we have

$2a=\sqrt{(7-6)^2+(0-2)^2}+\sqrt{(-7-6)^2+(0-2)^2}\\ 2a=\sqrt{5}+\sqrt{173}\\$

2a = 15.74

a = 7.87

Put a = 7.87 in c² = a² – b² to find b.

b² = 62 – 49

b² = 13

As the ellipse lies on x-axis, the equation is of the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$.

So, the equation is, $\frac{x^2}{62}+\frac{y^2}{13}=1$.

问题 2. 求一个椭圆的方程，通过焦点 (±5√3, 0) 和点 (6, 4) 的原点。

解决方案：

Using the formula, we have

$2a=\sqrt{(5\sqrt{3}-6)^2+(0-4)^2}+\sqrt{(-5\sqrt{3}-6)^2+(0-4)^2}$

2a = 20

a = 10

Put a = 10 in c² = a² – b² to find b.

b² = 100 – 75

b² = 25

As the ellipse lies on x-axis, the equation is of the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$.

So, the equation is, $\frac{x^2}{100}+\frac{y^2}{25}=1$.

问题 3.用焦点 (0, ±5) 和短轴 (12, 0) 找到椭圆长轴的坐标。

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 4. 如果 a = 3, b = 5并且长轴平行于 x 轴，则求椭圆通过原点的方程。

解决方案：

The major axis is parallel to x-axis, so the ellipse lies on x-axis.

The equation is of the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$.

Here, a = 3, b = 5, h = 0 and k = 0.

So, the equation becomes,

$\frac{(x-0)^2}{3^2}+\frac{(y-0)^2}{5^2}=1$

$\frac{x^2}{9}+\frac{y^2}{25}=1$

问题 5. 如果 a = 13，b = 5，并且短轴平行于 x 轴，则求椭圆通过原点的方程。

解决方案：

The major axis is parallel to x-axis, so the ellipse lies on x-axis.

The equation is of the form $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$. 

Here a = 13 and b = 5, h = 0 and k = 0.

So, the equation becomes,

$\frac{(x-0)^2}{5^2}+\frac{(y-0)^2}{13^2}=1$

$\frac{x^2}{25}+\frac{y^2}{169}=1$