第11类RD Sharma解决方案–第22章直角坐标直角坐标系的简要说明-练习22.3

问题1.当轴通过点(a–c，b)转移到平行轴时，方程(x – a) ² +(y – b) ² = r ^{2会变成什么?}

解决方案：

We are given,

(x – a)² + (y – b)² = r²

Putting x =  X + a – c and y = Y + b, we get,

=> ((X + a – c) – a)² + ((Y + b ) – b)² = r²

=> (X – c)² + Y² = r²

=> X² + c² – 2cX + Y² = r²

=> X² + Y² – 2cX = r² – c²

Therefore the required equation is X² + Y² – 2cX = r² – c².

问题2.如果原点不旋转就移动到点(ab /(a–b)，0)，方程(a – b)(x ² + y ^{2)– 2abx = 0会变成什么?}

解决方案：

We are given,

(a – b) (x² + y²) – 2abx = 0

Putting x = X + [ab/(a–b)] and y = Y, we get,

=> $(a - b)\left[\left(X+\frac{ab}{a+b}\right)^2 + Y^2\right] - 2ab\left(X+\frac{ab}{a+b}\right) = 0$

=> $(a-b)\left[X^2+\left(\frac{ab}{a+b}\right)^2+2\frac{Xab}{a+b}+Y^2\right]-2abX-2\frac{(ab)^2}{a-b} = 0$

=> $(a-b)\left[\frac{X^2(a-b)^2+(ab)^2+2Xab(a-b)+Y^2(a-b)^2}{(a-b)^2}\right]-\frac{2abX(a-b)+2(ab)^2}{a-b}= 0$ 

=> $\frac{X^2(a-b)^2+(ab)^2+2Xab(a-b)+Y^2(a-b)^2}{a-b}=\frac{2abX(a-b)+2(ab)^2}{a-b}$

=> X² (a–b)² + (ab)² + 2abX (a–b)+Y² (a–b)² = 2abX (a–b)+2(ab)²

=> (a – b)² (X² + Y²) = a²b²

Therefore the required equation is (a – b)² (X² + Y²) = a²b².

问题3.找到将原点移动到点(1，1)时的下列方程式吗?

(i)x ² + xy – 3x – y + 2 = 0

解决方案：

We are given,

x² + xy – 3x – y + 2 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)² + (X + 1) (Y + 1) – 3(X + 1) – (Y + 1) + 2 = 0

=> X² + 1 + 2X + XY + X + Y + 1 – 3X – 3 – Y – 1 + 2 = 0

Therefore the required equation is X² + XY = 0.

(ii)x ² – y ² – 2x + 2y = 0

解决方案：

We are given,

x² – y² – 2x + 2y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)² – (Y + 1)² – 2(X + 1) + 2(Y + 1) = 0

=> X² + 1 + 2X – Y² – 1 – 2Y – 2X – 2 + 2Y + 2 = 0

=> X² – Y² = 0

Therefore the required equation is X² – Y² = 0.

(iii)xy – x – y + 1 = 0

解决方案：

We are given,

xy – x – y + 1 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) – (X + 1) – (Y + 1) + 1 = 0

=> XY + X + Y + 1 – X – 1 – Y – 1 + 1 = 0

=> XY = 0

Therefore the required equation is XY = 0.

(iv)xy – y ² – x + y = 0

解决方案：

We are given,

xy – y² – x + y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) – (Y + 1)² – (X + 1) + (Y + 1) = 0

=> XY + X + Y + 1 – Y² – 1 – 2Y – X – 1 + Y + 1 = 0

=> XY – Y² = 0

Therefore, the required equation is XY – Y² = 0.

问题4.将原点移动到哪一点，以便等式x ² + xy – 3x – y + 2 = 0不包含任何一阶项和常数项?

解决方案：

We are given,

x² + xy – 3x – y + 2 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)² + (X + a)(Y + b) – 3(X + a) – (Y + b) + 2 = 0

=> X² + a² + 2aX + XY + aY + bX + ab – 3X – 3a – Y – b + 2 = 0

=> X² + XY + X(2a + b – 3) + Y(a – 1) + a² + ab – 3a – b + 2 = 0

As our transformed equation has no first-degree term, we have,

2a + b – 3 = 0 and a – 1 = 0   

By solving these equations we have a = 1 and b = 1.

Therefore, the origin has been shifted to (1,1) from (0,0).

问题5.验证当原点移至点(–1)时，具有顶点(2、3)，(5、7)和(–3，–1)的三角形的面积在轴的平移下保持不变。 ，3)。

解决方案：

Here, L.H.S. = A₁ = Area of the triangle with vertices (2, 3), (5, 7) and (–3, –1)

= $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

= $\frac{1}{2}|2(7+1)+5(-1-3)-3(3-7)|$

= $\frac{1}{2}|16-20-12|$ 

= $\frac{8}{2}$

= 4 sq. units

As the origin shifted to point (–1, 3), the new coordinates of the triangle are:

(X₁, Y₁) = (2–1, 3+3) = (1, 6) 

(X₂, Y₂) = (5–1, 7+3) = (4, 10)

(X₃, Y₃) = (–3–1, –1+3) = (–4, 2) 

Now, R.H.S. = A₂ = Area of the triangle with vertices (1, 6), (4, 10) and (–4, 2)

= $\frac{1}{2}|X_1(Y_2-Y_3)+X_2(Y_3-Y_1)+X_3(Y_1-Y_2)|$

= $\frac{1}{2}|1(10-2)+4(2-6)-4(6-10)|$

= $\frac{1}{2}|8-16+16|$

= $\frac{8}{2}$

= 4 sq. units

Therefore, A₁ = A₂.

Hence, proved.

问题6.找到将原点移动到点(1，1)时的下列方程式吗?

(i)x ² + xy – 3y ² – y + 2 = 0

解决方案：

We are given,

x² + xy – 3y² – y + 2 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)² + (X + 1) (Y + 1) – 3(Y + 1)² – (Y + 1) + 2 = 0

=> X² + 1 + 2X + XY + X + Y + 1 – 3Y² – 3 – 6Y – Y – 1 + 2 = 0

=> X² – 3Y² + XY + 3X – 6Y = 0

Therefore, the required equation is X² – 3Y² + XY + 3X – 6Y = 0.

(ii) xy – y ² – x + y = 0

解决方案：

We are given,

xy – y² – x + y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) – (Y + 1)² – (X + 1)+ Y + 1 = 0

=> XY + X + Y + 1 – Y² – 1 – 2Y – X – 1 + Y + 1 = 0

=> XY – Y² = 0

Therefore, the required equation is XY – Y² = 0.

(iii)xy – x – y + 1 = 0

解决方案：

We are given,

xy – x – y + 1 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) – (Y + 1) – (X + 1) + 1 = 0

=> XY + X + Y + 1 – Y – 1 – X – 1 + 1 = 0

=> XY = 0

Therefore, the required equation is XY = 0.

(iv)x ² – y ² – 2x + 2y = 0

解决方案：

We are given,

x² – y² – 2x + 2y = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)² – (Y + 1)² – 2(X + 1) + 2(Y + 1) = 0

=> X² + 1 + 2X – Y² – 1 – 2Y – 2X – 2 + 2Y + 2 = 0

=> X² – Y² = 0

Therefore, the required equation is X² – Y² = 0.

问题7：在轴平移后找到原点应移至的点，以便以下方程式不包含一阶项。

(i)x ² + y ² – 4x – 8y + 3 = 0

解决方案：

We are given,

x² + y² – 4x – 8y + 3 = 0 

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)² + (Y + b)² – 4(X + a) – 8(Y + b) + 3 = 0

=> X² + a² + 2aX + Y² + b² + 2bY – 4X – 4a – 8Y – 8b + 3 = 0

=> X² + Y² + (2a – 4)X + (2b – 8)Y + (a² + b² – 4a – 8b +3) = 0

As our transformed equation has no first-degree term, we have,

2a – 4 = 0 and 2b – 8 = 0 

By solving these equations we have a = 2 and b = 4.

Therefore, the origin has been shifted to (2,4) from (0,0).

(ii)x ² + y ² – 5x + 2y – 5 = 0

解决方案：

We are given,

x² + y² – 5x + 2y – 5 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)² + (Y + b)² – 5(X + a) + 2(Y + b) – 5 = 0

=> X² + a² + 2aX + Y² + b² + 2bY – 5X – 5a + 2Y + 2b – 5 = 0

=> X² + Y² + (2a – 5)X + (2b + 2)Y + (a² + b² – 5a + 2b – 5) = 0

As our transformed equation has no first-degree term, we have,

2a – 5 = 0 and 2b + 2 = 0  

By solving these equations we have a = 5/2 and b = –1.

Therefore, the origin has been shifted to (5/2, –1) from (0,0).

(iii)x ² – 12x + 4 = 0

解决方案：

We are given,

x² – 12x + 4 = 0

Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)² – 12(X + a) + 4 = 0

=> X² + a² + 2aX – 12X – 12a + 4 = 0

=> X² + (2a – 12)X + (a² – 12a + 4) = 0

As our transformed equation has no first-degree term, we have,

=> 2a – 12 = 0

=> a = 6

Therefore, the origin has been shifted to (6,b) from (0,0) where b is any arbitrary value.

问题8.验证当原点移至点(–2，2)时，具有顶点(4，6)，(7，10)和(1，–2)的三角形的面积在轴的平移下保持不变。 1)。

解决方案：

Here, L.H.S. = A₁ = Area of the triangle with vertices (4, 6), (7, 10) and (1, –2)

= $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

= $\frac{1}{2}|4(10+2)+7(-2-6)+1(6-10)|$

= $\frac{1}{2}|48-56-4|$

= $\frac{12}{2}$

= 6 sq. units

As the origin shifted to point (–2, 1), the new coordinates of the triangle are:

(X₁, Y₁) = (4–2, 6+1) = (2, 7)

(X₂, Y₂) = (7–2, 10+1) = (5, 11)

(X₃, Y₃) = (1–2, –2+1) = (–1, –1)

Now, R.H.S. = A₂ = Area of the triangle with vertices (2, 7), (5, 11), (–1, –1)

= $\frac{1}{2}|X_1(Y_2-Y_3)+X_2(Y_3-Y_1)+X_3(Y_1-Y_2)|$

= $\frac{1}{2}|2(11+1)+5(-1-7)-1(7-11)|$

= $\frac{1}{2}|24-40+4|$

= $\frac{12}{2}$

= 6 sq. units

Therefore, A₁ = A₂.

Hence, proved.