第11类RD Sharma解法–第22章，直角坐标直角坐标系的简要说明-练习22.2

问题1.找到与点(2，4)和y轴等距的点的轨迹。

解决方案：

Let C (a, b) be any point on the locus and let A (2, 4) and B (0, b). We are given,

=> CA = CB

=> CA² = CB²

Using distance formula, we get,

=> (a − 2)² + (b − 4)² = (a − 0)² + (b − b)²

=> a² + 4 − 4a + b² + 16 − 8b = a²

=> b² − 4a − 8b + 20 = 0

Replacing (a, b) with (x, y), we get the locus of our point,

=> y² − 4x − 8y + 20 = 0

Therefore, locus of the point is y² − 4x − 8y + 20 = 0.

问题2。找到一个点的轨迹方程，使其移动到(2，0)与(1，3)的距离之比为5：4。

解决方案：

Let C (a, b) be any point on the locus and let A (2, 0) and B (1, 3). We are given,

=> CA/CB = 5/4 

=> CA²/CB² = 25/16

Using distance formula, we get,

=> $\frac{(a-2)^2+(b-0)^2}{(a-1)^2+(b-3)^2}=\frac{25}{16}$ 

=> $\frac{a^2+4-4a+b^2}{a^2+1-2a+b^2+9-6b}=\frac{25}{16}$

=> 16 (a² + 4 − 4a + b²) = 25 (a² + 1 − 2a + b² + 9 − 6b)

=> 9a² + 9b² + 14a − 150b + 186 = 0

Replacing (a, b) with (x, y), we get the equation of the locus of our point,

=> 9x² + 9y² + 14x − 150y + 186 = 0

Therefore the locus of the point is 9x² + 9y² + 14x − 150y + 186 = 0. 

问题3。一个点移动，使其与(ae，0)和(-ae，0)的距离之差为2a，证明该点的等式为x ² / a ² -y ² / b ² = 1，其中b ² = ^{2(E 2} – 1)。

解决方案：

Let C (h, k) be any point on the locus and let A (ae, 0) and B (−ae, 0). We are given,

=> CA − CB = 2a 

Using distance formula, we get,

=> $\sqrt{(h-ae)^2+(k-0)^2}-\sqrt{(h-(-ae))^2+(k-0)^2}=2a$

=> $\sqrt{(h-ae)^2+(k-0)^2}=2a+\sqrt{(h-(-ae))^2+(k-0)^2}$

Squaring both sides, we get,

=> $(h-ae)^2+(k-0)^2=(2a+\sqrt{(h-(-ae))^2+(k-0)^2})^2$

=> h² + a²e² − 2aeh + k² = 4a² + (h + ae)² + k² + $4a\sqrt{(h+ae)^2+(k-0)^2}$

=> h² + a²e² − 2aeh + k² = 4a² + h² + a²e² + 2aeh + k² + $4a\sqrt{(h+ae)^2+(k-0)^2}$

=> −4aeh − 4a² = $4a\sqrt{(h+ae)^2+(k-0)^2}$

Squaring both sides again, we get,

=> −(eh + a) = (h + ae)² + k²

=> e²h² + a² + 2aeh = h² + a²e² + 2aeh + k²

=> h² (e² – 1) – k² = a² (e² – 1)

=> $\frac{h^2}{a^2}-\frac{k^2}{a^2(e^2-1)}=1$

As we are given, b² = a² (e² − 1), we get,

=> h²/a² − k²/b² = 1

Replacing (h, k) with (x, y), we get the equation of the locus of our point,

=> x²/a² − y²/b² = 1

Hence proved.

问题4.找到一个点的轨迹，使其与(0，2)和(0，-2)的距离之和为6。

解决方案：

Let C (a, b) be any point on the locus and let A (0, 2) and B (0, −2). We are given,

=> CA + CB = 6

Using distance formula, we get,

=> $\sqrt{(a-0)^2+(b-2)^2}+\sqrt{(a-0)^2+(b-(-2))^2}=6$

=> $\sqrt{(a-0)^2+(b-2)^2}=6-\sqrt{a^2+(b+2)^2}$

Squaring both sides, we get,

=> a² + b² + 4 − 4b= 36 + a² + b² + 4 + 4b − $12\sqrt{a^2+(b+2)^2}$

=> −8b − 36 = $-12\sqrt{a^2+(b+2)^2}$

=> −4 (2b + 9) = $-12\sqrt{a^2+(b+2)^2}$ 

Squaring both sides again, we get,

=> (2b + 9)² = $(3\sqrt{a^2+(b+2)^2})^2$ 

=> 4b² + 81 + 36b = 9a² + 9b² + 36b + 36

=> 9a² + 5b² = 45

Replacing (a, b) with (x, y), we get the locus of our point,

=> 9x² + 5y² = 45

Therefore the locus of the point is 9x² + 5y² = 45.

问题5.找到与(1，3)和x轴等距的点的轨迹。

解决方案：

Let C (a, b) be any point on the locus and let A (1, 3) and B (a, 0). We are given,

=> CA = CB

=> CA² = CB²

Using distance formula, we get,

=> (a − 1)² + (b − 3)² = (a − a)² + (b − 0)²

=> a² + 1 − 2a + b² + 9 − 6b = b²

=> a² − 2a − 6b + 10 = 0

Replacing (a, b) with (x, y), we get the locus of our point,

=> x² − 2x − 6y + 10 = 0

Therefore the locus of the point is x² − 2x − 6y + 10 = 0.

问题6.找到一个点的轨迹，使其与原点的距离为与x轴的距离的三倍。

解决方案：

Let C (a, b) be any point on the locus and let A (0, 0) and B (a, 0). We are given,

=> CA = 3 CB

=> CA² = 9 CB²

Using distance formula, we get,

=> (a − 0)² + (b − 0)² = 9 [(a − a)² + (b − 0)²]

=> a² + b² = 9b²

=> a² = 8b²

Replacing (a, b) with (x, y), we get the locus of our point,

=> x² = 8y²

Therefore the locus of the point is x² = 8y².

问题7. A(5，3)，B(3，-2)是两个固定点，找到点P的轨迹的方程式，点P的运动使得三角形PAB的面积为9平方单位。

解决方案：

Let P (a, b) be any point on the locus and we have A (5, 3) and B (3, −2). We are given,

=> Area of the triangle PAB = 9 

=> $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=9$

=> |5(−2−b) + 3(b−3) + h(3+2)| = 18

=> |5a − 2b − 19| = 18

=> 5a − 2b − 19 = ±18

=> 5a − 2b − 37 = 0 or 5a − 2b − 1 = 0

Replacing (a, b) with (x, y), we get the locus of our point,

=> 5x − 2y − 37 = 0 or 5x − 2y − 1 = 0

Therefore the equation to the locus of the point is 5x − 2y − 37 = 0 or 5x − 2y − 1 = 0.

问题8.找到一个点的轨迹，使具有端点(2，0)和(-2，0)的线段在该点处成直角。

解决方案：

Let C (a, b) be any point on the locus and let A (2, 0) and B (−2, 0). 

We are given ∠ ACB = 90^o

=> AB² = CA² + CB²

Using distance formula, we get,

=> (2+2)² + (0−0)² = (a−2)² + (b−0)² + (a+2)² + (b−0)²

=> 16 = a² + 4 − 4a + b² + a² + 4 + 4a + b²

=> 2a² + 2b² + 8 = 16

=> a² + b² = 4

Replacing (a, b) with (x, y), we get the locus of our point,

=> x² + y² = 4

Therefore the locus of the point is x² + y² = 4.

问题9. A(-1，1)，B(2，3)是两个固定点，找到一个点P的轨迹，点P的移动使得三角形PAB的面积为8平方单位。

解决方案：

Let P (a, b) be any point on the locus and we have A (−1, 1) and B (2, 3). We are given,

=> Area of the triangle PAB = 8

=> $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=8$

=> |−1(3−b) + 2(b−1) + a(1−3)| = 16

=> |−2a + 3b − 5| = 16

=> −2a + 3b − 5 = ±16

=> 2a − 3b + 21 = 0 or 2a − 3b − 11 = 0

Replacing (a, b) with (x, y), we get the locus of our point,

=> 2x − 3y + 21 = 0 or 2x − 3y − 11 = 0

Therefore the locus of the point is 2x − 3y + 21 = 0 or 2x − 3y − 11 = 0.

问题10.长度为l的杆在两条垂直线之间滑动。在杆上找到将其按1：2比例划分的点的轨迹。

解决方案：

Let C (h, k) be any point on the locus and let AB = l (given) be the length of the rod. Suppose, coordinates of A and B are (a, 0) and (0, b) respectively.

According to the question,

=> h = 2a/3

=> a = 3h/2  . . . . (1)

And k = b/3

=> b = 3k  . . . . (2)

Let the origin be O (0, 0). Now we know △ AOB is right-angled.

=> AB² = OA² + OB²

=> l² = [(a−0)² + (0−0)²] + [(0−0)² + (b−0)²]

=> a² + b² = l²

Using (1) and (2), we get, 

=> (3h/2)² + (3k)² = l²

=> 9h²/4 + 9k² = l²

=> 9h² + 36k² = 4l²

Replacing (h, k) with (x, y), we get the locus of our point,

=> 9x² + 36y² = 4l²

Therefore the locus of the point is 9x² + 36y² = 4l².

问题11。找到线x cosα+ y sinα= p的部分的中点的轨迹，该点在轴之间截取。

解决方案：

We are given,

=> x cos α + y sin α = p

=> $\frac{x}{\frac{p}{cos α}}+\frac{y}{\frac{p}{sin α}}=1$

Intercepts on x-axis and y -axis are p/cos α and p/sin α respectively.

Suppose (x, y) is the mid-point of the portion of the given line which is intercepted between the axes.

=> (x, y) = $\left(\frac{\frac{p}{cosα}+0}{2},\frac{\frac{p}{sinα}+0}{2}\right)=\left(\frac{p}{2cosα},\frac{p}{2sinα}\right)$ 

=> x = p/2 cos α and y = p/2 sin α

=> 2 cos α = p/x and 2 sin α = p/y

Squaring both sides of these, we get,

=> 4 cos² α = p²/x²  . . . . (1)

=> 4 sin² α = p²/y²  . . . . (2)

Adding (1) and (2), we get,

=> 4 cos² α + 4 sin² α = p²/x² + p²/y² 

=> p²/x² + p²/y² = 4

=> p² (x² + y²) = 4x²y²

Therefore the locus of the mid-point is p² (x² + y²) = 4x²y².

问题12.如果O是原点，而Q是y ² = x上的变量点。找到OQ中点的位置。

解决方案：

Let P (h, k) be the point on the locus and let Q (a, b). 

According to the question,

=> h = (a+0)/2 and k = (b+0)/2

=> h = a/2 and k = b/2

=> a = 2h and b = 2k

As point Q lies on y² = x, we get,

=> (2k)² = 2h

=> 4k² = 2h

=> 2k² = h

Replacing (h, k) with (x, y), we get the locus of our point,

=> 2y² = x

Therefore the locus of the mid-point is 2y² = x.