# Equation of Locus After Shifting Origin

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### Type – I: To Find New Equation of Locus After Shifting the Origin

#### Example – 01:

• The origin is shifted to the point (1, -4). Find the new equation of locus 2x2 – xy + 3y2 – 8x + 25 y + 54 = 0 axes remaining parallel.
• Solution:

The old equation of locus is 2x2 – xy + 3y2 – 8x + 25 y + 54 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, -4)

We have x = X + h = X + 1 and y = Y + k = Y – 4

Substituting these values in equation (1) we have

2(X + 1)2 – (X + 1)( Y – 4) + 3( Y – 4)2 – 8(X + 1) + 25 ( Y – 4) + 54 = 0

∴  2(X2 + 2X + 1) – (XY – 4X + Y – 4) + 3( Y2 – 8Y + 16) – 8X -8 + 25 Y – 100 + 54 = 0

∴  2X2 + 4X + 2 – XY + 4X – Y + 4 + 3Y2 – 24Y + 48 – 8X -8 + 25 Y – 100 + 54 = 0

∴  2X2 – XY + 3Y2 = 0

Ans: The new equation of locus is 2X2 – XY + 3Y2 = 0

#### Example – 02:

• The origin is shifted to the point (1, 1). Find the new equation of locus xy – x – y + 1 = 0 axes remaining parallel.
• Solution:

The old equation of locus is xy – x – y + 1 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, 1)

We have x = X + h = X + 1 and y = Y + k = Y + 1

Substituting these values in equation (1) we have

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

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

∴  XY  = 0

Ans: The new equation of locus is XY  = 0

#### Example – 03:

• The origin is shifted to the point (1, 1). Find the new equation of locus x2 – y2 – 2x + 2y = 0 axes remaining parallel.
• Solution:

The old equation of locus is x2 – y2 – 2x + 2y = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, 1)

We have x = X + h = X + 1 and y = Y + k = Y + 1

Substituting these values in equation (1) we have

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

∴  (X2 + 2X + 1) – ( Y2 + 2Y + 1) – 2X -2 + 2Y  + 2 = 0

∴  X2 + 2X + 1 – Y2 – 2Y  – 1 – 2X -2 + 2Y  + 2 = 0

∴  X2 – Y2 = 0

Ans: The new equation of locus is X2 – Y2 = 0

#### Example – 04:

• The origin is shifted to the point (1, 1). Find the new equation of locus x2 + y2 – 4x + 6y +3 = 0 axes remaining parallel.
• Solution:

The old equation of locus is x2 + y2 – 4x + 6y +3 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, 1)

We have x = X + h = X + 1 and y = Y + k = Y + 1

Substituting these values in equation (1) we have

(X + 1)2 + (Y + 1)2 – 4(X + 1) + 6(Y + 1) + 3 = 0

∴  (X2 + 2X + 1) + (Y2 + 2Y + 1) – 4X – 4 + 6Y + 6 + 3 = 0

∴  X2 + 2X + 1 + Y2 + 2Y + 1 – 4X – 4 + 6Y + 6 + 3 = 0

∴  X2 + Y2 – 2X + 8Y + 7 = 0

Ans: The new equation of locus is X2 + Y2 – 2X + 8Y + 7 = 0

#### Example – 05:

• The origin is shifted to the point (2, -1). Find the new equation of locus 2x2 + 3xy – 9y2 – 5x – 24y – 7 = 0 axes remaining parallel.
• Solution:

The old equation of locus is 2x2 + 3xy – 9y2 – 5x – 24y – 7 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (2, -1)

We have x = X + h = X + 2 and y = Y + k = Y – 1

Substituting these values in equation (1) we have

2(X + 2)2 + 3(X + 2) (Y – 1) – 9(Y – 1)2 – 5(X + 2) – 24(Y – 1) – 7 = 0

∴  2(X2 + 4X + 4) + 3(XY – X + 2Y – 2) – 9(Y2 – 2Y + 1) – 5X – 10 – 24Y + 24 – 7 = 0

∴  2X2 + 8X + 8 + 3XY – 3X + 6Y – 6 – 9Y2 + 18Y  – 9 – 5X – 10 – 24Y + 24 – 7 = 0

∴  2X2 + 3XY – 9Y2 = 0

Ans: The new equation of locus is 2X2 + 3XY – 9Y2 = 0

#### Example – 06:

• The origin is shifted to the point (a – c, b). Find the new equation of locus (x – a)2 + (y – b)2 = r2 axes remaining parallel.
• Solution:

The old equation of locus is (x – a)2 + (y – b)2 = r2 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (a – c, b)

We have x = X + h = X + a – c and y = Y + k = Y + b

Substituting these values in equation (1) we have

(X + a – c – a)2 + (Y + b – b)2 = r2

∴  (X – c)2 + (Y)2 = r2

∴  (X – c)2 + (Y)2 = r2

∴  X2 – 2cX + c2 + Y2 = r2

∴  X2 + Y2 – 2cX + c2 – r2 = 0

Ans: The new equation of locus is X2 + Y2 – 2cX + c2 – r2 = 0

#### Example – 07:

• The origin is shifted to the point (1/2, -3/2). Find the new equation of locus (2x – 1)2 + (y + 3/2)2 = 4 axes remaining parallel.
• Solution:

The old equation of locus is (2x – 1)2 + (y + 3/2)2 = 4  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1/2, -3/2)

We have x = X + h = X + 1/2 and y = Y + k = Y – 3/2

Substituting these values in equation (1) we have

(2(X + 1/2) – 1)2 + ((Y – 3/2) + 3/2)2 = 4

∴  (2X + 1 – 1)2 + (Y – 3/2 + 3/2)2 = 4

∴  (2X)2 + (Y)2 = 4

∴  4X2 + Y2 = 4

Ans: The new equation of locus is 4X2 + Y2 = 4

#### Example – 08:

• The origin is shifted to the point (-1, 2). Find the new equation of locus 4x2 + y2 + 6x – 4y + 5 = 0 axes remaining parallel.
• Solution:

The old equation of locus is 4x2 + y2 + 8x – 4y + 4 = 0  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (2, -1)

We have x = X + h = X – 1 and y = Y + k = Y + 2

Substituting these values in equation (1) we have

4(X – 1)2 + (Y + 2)2 + 8(X – 1) – 4(Y + 2) + 4 = 0

∴  4(X2 – 2X +  1) + (Y2 + 4Y + 4) + 8X – 8 – 4Y – 8 + 4 = 0

∴  4X2 – 8X +  4 + Y2 + 4Y + 4 + 8X – 8 – 4Y – 8 + 4 = 0

∴  4X2 + Y2 – 4 = 0

∴  4X2 + Y2 = 4

Ans: The new equation of locus is 4X2 + Y2 = 4

#### Example – 09:

• Obtain the new equation of locus (a – b)(x2 + y2) – 2abx = 0. If the origin is shifted to the point (ab/(a-b), 0) axes remaining parallel.
• Solution:

The old equation of locus is (a – b)(x2 + y2) – 2abx = 0   …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (2, -1)

We have x = X + h = X + ab/(a-b) and y = Y + k = Y + 0 = Y

Substituting these values in equation (1) we have

Ans: The new equation of locus is (a – b)2(X2 + Y2) – a2b= 0

### Type – II: To Find Old Equation of Locus After Shifting the Origin

#### Example – 10:

• The origin is shifted to the point (5, -3). The equation of locus in new system is Y2 = 6X. Find the equation of locus in the old system
• Solution:

The new equation of locus is Y2 = 6X  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (5, -3)

We have X = x – h = x – 5 and Y = y – k = y + 3

Substituting these values in equation (1) we have

(y + 3)2 = 6(x – 5)

∴  y2 + 6y + 9 = 6x – 30

∴  y2 – 6x + 6y + 39 = 0

Ans: The equation of locus in old system is y2 – 6x + 6y + 39 = 0

#### Example – 11:

• The origin is shifted to the point (-1, 2). The equation of locus in new system is X2 + 5XY + 3Y2 = 0. Find the equation of locus in the old system
• Solution:

The new equation of locus is X2 + 5XY + 3Y2 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (-1, 2)

We have X = x – h = x + 1 and Y = y – k = y – 2

Substituting these values in equation (1) we have

(x + 1)2 + 5(x + 1)(y -2) + 3(y -2)2 = 0

∴  (x2 + 2x + 1) + 5(xy – 2x + y – 2) + 3(y2 – 4y + 4) = 0

∴  x2 + 2x + 1 + 5xy – 10x + 5y – 10 + 3y2 – 12y + 12 = 0

∴  x2 + 5xy + 3y2 – 8x – 7y + 3 = 0

Ans: The equation of locus in old system is x2 + 5xy + 3y2 – 8x – 7y + 3 = 0

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