圆锥截面是当平面以不同角度与右圆锥相交时生成的曲线。圆锥截面的名称中具有定义,它是通过切割圆锥体而形成的截面。每种类型的交点都提供不同类型的曲线。根据平面相交的角度,曲线的形状会发生变化,并形成不同的曲线,例如抛物线,双曲线,椭圆形等。
我们可以从相交处获得以下曲线:
- 圆圈
- 抛物线
- 双曲线
- 椭圆
每条曲线的方程式都有某些共同的特征。让我们研究它们中的每一个,看看如何从它们的方程式中识别它们。
圆圈
Circle is set of points which are equidistant from a fixed point.
下图表示一个圆,其圆心由O给出,半径是将圆心连接到圆上任何点的线。
其等式由下式给出:
(x – h) 2 +(y – k) 2 = r 2
其中(h,k)是圆的中心,并且半径由“ r”给出。可以重新排列此等式以使其具有以下形式:
x 2 + y 2 -2hx -2kx + h 2 + k 2 = r 2
从方程中识别圆
一个圆在其方程式中将同时具有x和y平方,且系数均为非零。 x 2和y 2必须具有相同的符号。一旦我们知道它是圆的方程,就可以恢复曲线。
例如:
The equation given is, 4x2+ 4y2+ 7y= 9, The easiest way to identify when the given equation is the equation of a circle,
1. Both squares of x and y are present in the equation.
2. The coefficients of the squares of x and y are same (here, +4).
These two information tells that the provided curve’s equation is a Circle.
抛物线
A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane
固定线称为Directrix,点称为焦点。
抛物线的标准方程式
以x轴为轴,经过(a,0)的原点和焦点的标准抛物线方程,由下式给出:
y 2 = 4ax
下图显示了不同类型的抛物线及其方程式。
从方程式中识别抛物线
从展开方程中识别抛物线方程。我们只需要注意一件事。抛物线方程应具有y或x平方,但不能同时具有两者。其余的值可以是任何值。如果方程式有x或y平方,但不是两个都平方。然后可以以抛物线的标准方程式的形式对其进行重新排列。
例如:
Take a look at these equations,
x2= y+ 4, y2– 3x+ 9= 0
Both the equations mentioned above are a Parabola because,
1. Both equations have one of the variables squared, but not both. Therefore, the equations are parabola.
椭圆
An ellipse is a set of all the points in a plane, the sum of whose distances from two fixed points in a plane is constant.
上面定义中提到的两点称为椭圆的焦点。
以x轴为长轴,y轴为短轴的椭圆的标准方程由下式给出:
其中,c 2 = a 2 – b 2
如果椭圆的长轴在y轴上,依此类推。然后,等式由下式给出:
从方程式中识别椭圆:与圆形相似,椭圆也具有x和y正方形。但是不同之处在于它们将具有不同的系数。
例如:
5x2+ 7y2– 9x- 6y = 0, 9y2+ 2y2+ 8x+ y= 0
Both the equations are the curves of ellipse because,
1. Both x and y variables are squared.
2. The coefficients of both squared variables are different in values (if the coefficients were equal, the curve would be a Circle)
双曲线
Hyperbola is a set of points in a plane, the difference of whose distances from two fixed points in the plane is a constant.
双曲线的标准方程式为
同样在这里,c 2 = a 2 – b 2
在上述情况下,横轴是x轴,共轭轴是y轴。如果轴反转,则方程将变为
从方程式中识别双曲线:与圆相似,双曲线也具有x和y平方。但是不同之处在于它们将具有不同的系数,并且符号将相反。
例如:
5x2– 2y2+ 7x= 8
The above equation is the equation of a Hyperbola since,
1. Both x and y with a degree 2 present.
2. one of the squared variables have different sign than other, in this case y2 coefficient is negative and x2 coefficient is positive.
让我们看一下有关这些概念的一些示例问题
样本问题
问题1:从扩展方程中识别曲线。
y 2 -4y + 2 = 12x
解决方案:
y2 – 4y + 2 = 12x
This equation contains only y square. We have seen in the previous sections that the parabola equations have either x or y squared. So, this must be the equation of the parabola.
Rearranging the given equation,
y2 – 4y + 2 = 12x
⇒ y2 – 4y + 4 + 2 – 4 = 12x
⇒(y – 2)2 – 2 =12x
⇒(y – 2)2 = 12x + 2
⇒
问题2:从扩展方程中识别曲线:
4x 2 + 9y 2 = 36
解决方案:
4x2 + 9y2 = 36
The given equation has both x and y squares present in it. Both have positive but different coefficients. So, it might be an equation of ellipse.
So, this is the equation with a = 3 and b = 2.
问题3:根据扩展方程确定曲线:
7x 2 – 9y 2 = 36
解决方案:
7x2 – 9y2 = 36
This equation also has squares of both x and y, but the signs are different. Based on the above-mentioned method, we can say that this is a hyperbola.
Now we need to bring the expanded equation in standard form.
Here, a = and b = 2
问题4:给定曲线的扩展方程式,将其识别出来,然后将其恢复为标准形式。
x 2 + y 2 + 6y = 27
解决方案:
x2 + y2 + 6y = 27
Both x and y squares are present and have same sign and 1 as their coefficient. This is an equation of a circle.
x2 + y2 + 6y = 27
⇒ x2 + y2 + 6y + 9 = 27 + 9
⇒ x2 + (y + 3)2 = 36
⇒ x2 + (y + 3)2 = 62
This is the equation of circle with centre at (0,-3) and radius 6.
问题5:根据给定的表达式确定曲线并公式化方程。
x 2 + y 2 + 4x + 6x = 12
解决方案:
Let’s take the given equation,
x2 + y2 + 4x + 6x = 12
From the equations we have studied above, notice that in the given equation x and y both are squared and both have the same sign and same coefficients. As mentioned previously, equations of circle should have both x and y squares with same sign and coefficients. Thus, this is the equation of the circle.
To find the centre and the radius of the circle, we need to rearrange it.
x2 + y2 + 4x + 6x = 12
We need to make whole squares out of the x and y terms
x2 + 4x + y2 + 6x = 12
⇒ x2 + 4x + 4 – 4 + y2 + 6x + 9 – 9 = 12
⇒ (x + 2)2 -4 + (y + 3)2 – 9 = 12
⇒ (x + 2)2 + (y + 3)2 = 12 + 9 + 4
⇒(x + 2)2 + (y + 3)2 = 25
⇒(x + 2)2 + (y + 3)2 = 52
So, the centre is (-2,-3) and the radius is 5.