将 r sin θ = 4 转换为矩形形式
圆锥截面是数学的一个分支,它处理由圆锥与平面相交形成的曲线的研究。圆锥和平面之间的相交可以根据相交的点和角度而有所不同。在考虑圆锥曲线方程时,最重要的两个是极坐标方程和矩形方程。在本文中,我们将讨论极坐标和矩形方程以及它们的转换。
极坐标方程
涉及角度θ和距离“r”之间关系的方程属于这一类。这里,“r”表示原点(也称为极点)和曲线上一点之间的距离, θ表示原点、x 轴的正侧和给定数学曲线上的一点所形成的顺时针角度.
x = r cos θ
y = r sin θ
矩形方程
涉及变量的方程属于此类矩形方程。这些方程中不涉及角度,它们只是具有适当数学运算符的常数和变量。矩形方程可以很容易地表示为笛卡尔平面上的图形。
r2 = x2 + y2
将极坐标方程转换为矩形方程
极坐标到矩形形式的转换很容易。要将极坐标转换为矩形,请执行以下步骤:
第 1 步:列出具有 r、 θ角和变量 x、y 的极坐标方程
x = r cos θ (eq1)
y = r sin θ (eq2)
第 2 步:仔细考虑可以消除θ角的情况。
对 eq1 求平方,我们得到:
x 2 = r 2 cos 2 θ (eq3)
同样,对 eq 2 求平方,我们得到:
y 2 = r 2 sin 2 θ (eq4)
第 3 步:根据涉及“Sin theta”和“Cos theta”的三角方程,根据需要执行数学加法
添加 eq3 和 eq4
x 2 + y 2 = r 2 cos 2 θ + r 2 cos 2 θ
我们已经知道, sin 2 θ + cos 2 θ = 1
r 2 = x 2 + y 2
我们看到从组合方程中消除了 theta,因此通过上述方式,我们能够将 Polar 方程转换为矩形方程。
将极坐标方程 r sin θ = 4 转换为矩形形式
解决方案:
Given r sin θ = 4 …(1)
As we know that
y = r sin θ
So, in the equation (1) replace r sinθ by y
y = 4
We conclude that the rectangular form of r sinθ = 4 is y = 4.
示例问题
问题 1. 将极坐标方程 r = 10 sin θ转换为矩形形式
解决方案:
Given r = 4 sinθ
Multiply both LHS and RHS by r
r2 = 4r sinθ
We already know, r2 = x2 + y2
Replace r2 by x2 + y2
x2 + y2 = 4r sinθ
Now, as we know y = r sinθ
So, replace r sinθ by y
x2 + y2 = 4 y
x2 + y2 – 4y = 0
Add 4 on both LHS and RHS
x2 + y2 + 4 – 4y = 4
x2 + (y – 2)2 = 4
We conclude that the rectangular form of r = 4 sin theta is x2 + (y – 2)2 = 4
问题 2. 将极坐标方程 r = 6 sin θ转换为矩形形式
解决方案:
Given r = 6 sinθ
Multiply both LHS and RHS by r
r2 = 6r sinθ
We already know, r2 = x2 + y2
Replace r2 by x2 + y2
x2 + y2 = 6r sinθ
Now, as we know y = r sinθ
So, replace r sinθ by y
x2 + y2 = 6y
x2 + y2 – 6y = 0
Add 9 on both LHS and RHS
x2 + y2 + 9 – 6y = 9
x2 + (y – 3)2 = 9
We conclude that the rectangular form of r = 6 sin theta is x2 + (y – 2)2 = 9
问题 3. 将极坐标方程 r = 4 cos θ转换为矩形形式
解决方案:
Given r = 4 cos theta
Multiply both LHS and RHS by r
r2 = 4 r cosθ
We already know, r2 = x2 + y2
Replace r2 by x2 + y2
x2 + y2 = 4r cosθ
Now, as we know x = r cosθ
So, replace r cosθ by y
x2 + y2 = 4 x
x2 + y2 – 4x = 0
Add 4 on both LHS and RHS
x2 + y2 + 4 – 4x = 4
(x – 2)2 + y2 = 4
We conclude that the rectangular form of r = 4 cos theta is (x – 2)2 + y2 = 4
问题 4. 给定方程 r = 6 cos θ,现在将其转换为矩形形式
解决方案:
Given r = 6 cosθ
Multiply both LHS and RHS by r
r2 = 6 r cosθ
We already know, r2 = x2 + y2
Replace r2 by x2 + y2
x2 + y2 = 6r cosθ
Now, as we know x = r cosθ
So, replace r cosθ by y
x2 + y2 = 6 x
x2 + y2 – 6x = 0
Add 9 on both LHS and RHS
x2 + y2 + 9 – 6x = 9
(x – 3)2 + y2 = 9
We conclude that the rectangular form of r = 6 cos theta is (x – 3)2 + y2 = 9
问题 5. 将极坐标方程 r = 10 sin θ转换为矩形形式
解决方案:
Given r = 10 sinθ
Multiply both LHS and RHS by r
r2 = 10r sinθ
We already know, r2 = x2 + y2
Replace r2 by x2 + y2
x2 + y2 = 10r sinθ
Now, as we know y = r sinθ
So, replace r sinθ by y
x2 + y2 = 10 y
x2 + y2 – 10y = 0
Add 25 on both LHS and RHS
x2 + y2 + 25 – 10y = 25
x2 + (y – 5)2 = 25
We conclude that the rectangular form of r = 10 sin theta is x2 + (y – 5)2 = 25