如何找到给定中心和两个点的椭圆方程？

圆锥截面可以定义为描述直圆锥与平面的交点的一组点。平面与圆锥的交角决定了圆锥截面的形状。当这个角度是锐角时，即在 45° 和 90° 之间。

或者，椭圆也可以这样定义：由一组点形成的闭合曲线，这些点到两个固定点的距离之和为常数。

The standard equation of the ellipse whose center is (h, k) is,

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

问题：确定给定椭圆的标准形式方程：两点P(p,q)和M(m,n)的坐标，以及圆心坐标O(h,k)。

解决方案：

It is known that the standard equation of the ellipse is,

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

We have been provided the value of (h, k) in the question. Both P(p, q) and M(m, n) will satisfy the equation.

So, now we have two equations:

Equation 1: $\dfrac{(p-h)^2}{a^2} + \dfrac{(q-k)^2}{b^2} = 1$

Equation 2:  $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$

We can equate both the LHS of the equations (since the RHS is equal). Then, we get:

$\dfrac{(p-h)^2}{a^2} + \dfrac{(q-k)^2}{b^2} = \dfrac{(m-h)^2}{a^2} + \dfrac{(n-k)^2}{b^2}$

We can simplify this as:

$\dfrac{(p-h)^2 -(m-h)^2}{a^2} = \dfrac{(n-k)^2 - (q-k)^2}{b^2}$

Using the formula a² – b² = (a – b)×(a + b), we can simplify the equation further:

$\dfrac {(p - m)\times(p + m - 2h)}{a ^ 2} = \dfrac {(n - q)\times(n +q- 2k)}{b^ 2}$

From this, we can get the relation between a and b as:

$a = b \times \sqrt \dfrac {(p - m)\times(p + m - 2h)}{(n - q)\times(n +q- 2k)}$

Now, we can plug this value in Equation 1 or Equation 2 to solve it finally.

Let us plug it into equation 1:

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

$\dfrac{(p-h)^2}{ b^2 \times\dfrac {(p - m)×(p + m - 2h)}{(n - q)×(n +q- 2k)} } + \dfrac{(q-k)^2}{b^2} = 1$

Finally, we can deduce that b is:

$b = \sqrt {\dfrac{(p-h)^2 \times (n - q)\times(n +q- 2k)}{(p - m)\times(p + m - 2h)} + (q-k)^2 }$

And hence, a is:

$a = \sqrt {\dfrac{(p-h)^2 (n - q)(n +q- 2k)}{(p - m)(p + m - 2h)} + (q-k)^2 } \times \sqrt \dfrac {(p - m)(p + m - 2h)}{(n - q)(n +q- 2k)}$

Given two points P (p, q) and M (m, n) of an ellipse with center (h, k), the equation of the ellipse in standard form is: 

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

where,

$a = \sqrt {\dfrac{(p-h)^2 \times (n - q)\times(n +q- 2k)}{(p - m)\times(p + m - 2h)} + (q-k)^2 } \times \sqrt \dfrac {(p - m)\times(p + m - 2h)}{(n - q)\times(n +q- 2k)}$

and 

$b = \sqrt {\dfrac{(p-h)^2 \times (n - q)\times(n +q- 2k)}{(p - m)\times(p + m - 2h)} + (q-k)^2 }$

示例问题

问题1：假设椭圆的中心是(5, 2)，点A(3, 4)和B(5, 6)穿过椭圆，形成椭圆的标准方程。

解决方案：

Given that center of the ellipse is (h, k) = (5, 2) and (p, q) = (3, 4) and (m, n) = (5, 6) are two points on the ellipse. 

We can use the values of a and b from the above formula to create the standard equation as:

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

We know that,

$a = \sqrt {\dfrac{(p-h)^2 \times (n - q)\times(n +q- 2k)}{(p - m)\times(p + m - 2h)} + (q-k)^2 } \times \sqrt \dfrac {(p - m)\times(p + m - 2h)}{(n - q)\times(n +q- 2k)}$

Substituting the values of p, q, h, k, m and n we get:

$\sqrt {\dfrac{(3-5)^2 \times (6 - 4)\times(6 +4- 22)}{(3 - 5)×(3 + 5 - 25)} + (4-2)^2 } \times \sqrt \dfrac {(3 - 5)\times(3 + 5 - 25)}{(6 - 4)×(6 +4- 22)}$

Solving, we get a ≈ 2.31.

Similarly, substituting the values of p, q, h, k, m and n in the formula for b, we get:

$\sqrt {\dfrac {(3-5)^2 \times (6 - 4)\times(6 +4- 22)}{(3 - 5)\times(3 + 5 - 25)} + (4-2)^2 }$

Solving, we get b = 4.

Therefore, the equation of the ellipse is:

$\dfrac{(x-5)^2}{5.33} + \dfrac{(y-2)^2}{16} = 1$

问题 2：假设椭圆的中心在 (5, 8)。点 A (9, 2) 和 B (7, 6) 通过椭圆。这样的椭圆是否可能存在于真实平面中？解释你的答案。

解决方案：

Our hypothesis is that the ellipse exists. 

Given that center of the ellipse is (h, k) = (5, 8) and (p, q) = (9, 2) and (m, n) = (7, 6) are two points on the ellipse, we can use the values of a and b from the above formula to create the standard equation  

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

We know that,

$a = \sqrt {\dfrac{(p-h)^2 \times (n - q)\times(n +q- 2k)}{(p - m)\times(p + m - 2h)} + (q-k)^2 } \times \sqrt \dfrac {(p - m)\times(p + m - 2h)}{(n - q)\times(n +q- 2k)}$

Substituting the values of p, q, h, k, m and n we get:

$a = \sqrt {\dfrac{(9-5)^2 \times (6 - 2)\times(6 +2- 28)}{(9 - 7)\times(9 + 7 - 25)} + (2-8)^2 } \times \sqrt \dfrac {(9 - 7)\times(9 + 7 - 25)}{(6 - 2)\times(6 +2- 28)}$

We cannot solve this and get a real value of a. 

Therefore, our hypothesis is False. An ellipse with the given conditions cannot exist.

问题3：假设椭圆的中心是(1, 4)，点A(2, 9)和B(12, 5)通过椭圆，形成椭圆的标准方程。

解决方案：

Given that center of the ellipse is (h, k) = (1, 4) and (p, q) = (2, 9) and (m, n) = (12, 5) are two points on the ellipse, we can use the values of a and b from the above formula to create the standard equation:

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

We know that,

$a=\sqrt {\dfrac{(p-h)^2 \times (n - q)\times(n +q- 2k)}{(p - m)\times(p + m - 2h)} + (q-k)^2 } \times \sqrt \dfrac {(p - m)\times(p + m - 2h)}{(n - q)\times(n +q- 2k)}$

Substituting the values of p, q, h, k, m and n we get:

$a = \sqrt {\dfrac{(2-1)^2 \times (5 - 9)\times(5 +9- 24)}{(2 - 12)\times(2 + 12 - 21)} + (9-4)^2 } \times \sqrt \dfrac {(2 - 12)\times(2 + 12 - 21)}{(5 - 9)\times(5+9-24)}$

Solving, we get a ≈ 11.22.

Similarly, substituting the values of p, q, h, k, m and n in the formula for b, we get:

$b = \sqrt {\dfrac{(2-1)^2 (5 - 9)(5+9-24)}{(2 - 12)(2 + 12 - 21)} + (9-4)^2 }$

Solving, we get b ≈ 5.02.

Therefore, the equation of the ellipse is:

$\dfrac{(x-1)^2}{126} + \dfrac{(y-4)^2}{25.2} = 1$

问题 4：假设椭圆的中心在 (1, 3)。点 A (8, 2) 和 B (7, 5) 通过椭圆。这样的椭圆是否可能存在于真实平面中？解释你的答案。

解决方案：

Our hypothesis is that the ellipse exists. 

Given that center of the ellipse is (h, k) = (1, 3) and (p, q) = (8, 2) and (m, n) = (7, 5) are two points on the ellipse, we can use the values of a and b from the above formula to create the standard equation:

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

We know that, 

$a = \sqrt {\dfrac{(p-h)^2(n - q)(n +q- 2k)}{(p - m)(p + m - 2h)} + (q-k)^2 }\times \sqrt \dfrac {(p - m)(p + m - 2h)}{(n - q)(n +q- 2k)}$

Substituting the values of p, q, h, k, m and n we get:

$a = \sqrt {\dfrac{(8-1)^2(5 - 2)(5 +2- 23)}{(8 - 7)(8 + 7 - 21)} + (2-3)^2 }\sqrt \frac {(8 - 7)(8 + 7 - 21)}{(5 - 2)(5 +2- 23)}$

Solving this, we get a ≈ 7.3.

Similarly, substituting the values of p, q, h, k, m, and n in the formula for b, we get:

$b = \sqrt {\dfrac{(8-1)^2 (5 - 2)(5+2- 23)}{(8 - 7)(8+7-21)} + (2-3)^2 }$

Solving this, we get b  ≈ 3.51.

As the values of a and b are real, our hypothesis is True. An ellipse that satisfies the given conditions can exist.

问题 5：椭圆的中心在原点。点 (1, 5) 位于椭圆上。点 (11, 2) 是否也在椭圆上？

解决方案：

Our hypothesis is that an ellipse with a center at Origin (0, 0) passing through the points (11, 2) and (1, 5) exists. exists.

Now, we can use the values of a and b from the above formula to create the standard equation  

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

We know that,

 $a = \sqrt {\dfrac{(p-h)^2 (n - q)(n +q- 2k)}{(p - m)(p + m - 2h)} + (q-k)^2 } \sqrt \dfrac {(p - m)(p + m - 2h)}{(n - q)(n +q- 2k)}$

Substituting the values of p, q, h, k, m and n we get:

$a = \sqrt {\dfrac{(11-0)^2 (5 - 2) (5 +2- 20)}{(11 - 1) (11 + 1 - 20)} + (2-0)^2 } \times \sqrt \dfrac {(11 - 1)(11 + 1 - 20)}{(5 - 2)(5 +2- 20)}$

Solving this, we get a ≈ 11.99.

Similarly, substituting the values of p, q, h, k, m and n in the formula for b, we get:

$b = \sqrt {\dfrac{(11-0)^2 (5 - 2)(5 +2- 20)}{(11 - 1)(11 + 1 - 20)} + (2-0)^2 }$

Solving this, we get b  ≈ 5.02.

As the values of a and b are real, our hypothesis is True. The point (11, 2) does lie on the ellipse.