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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 2x
^{2}– xy + 3y^{2}– 8x + 25 y + 54 = 0 axes remaining parallel. **Solution:**

The old equation of locus is 2x^{2} – xy + 3y^{2} – 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(X^{2} + 2X + 1) – (XY – 4X + Y – 4) + 3( Y^{2} – 8Y + 16) – 8X -8 + 25 Y – 100 + 54 = 0

∴ 2X^{2} + 4X + 2 – XY + 4X – Y + 4 + 3Y^{2} – 24Y + 48 – 8X -8 + 25 Y – 100 + 54 = 0

∴ 2X^{2} – XY + 3Y^{2} = 0

**Ans:** The new equation of locus is 2X^{2} – XY + 3Y^{2} = 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 x
^{2}– y^{2}– 2x + 2y = 0 axes remaining parallel. **Solution:**

The old equation of locus is x^{2} – y^{2} – 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

∴ (X^{2} + 2X + 1) – ( Y^{2} + 2Y + 1) – 2X -2 + 2Y + 2 = 0

∴ X^{2} + 2X + 1 – Y^{2} – 2Y – 1 – 2X -2 + 2Y + 2 = 0

∴ X^{2} – Y^{2} = 0

**Ans:** The new equation of locus is X^{2} – Y^{2} = 0

#### Example – 04:

- The origin is shifted to the point (1, 1). Find the new equation of locus x
^{2}+ y^{2}– 4x + 6y +3 = 0 axes remaining parallel. **Solution:**

The old equation of locus is x^{2} + y^{2} – 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

∴ (X^{2} + 2X + 1) + (Y^{2} + 2Y + 1) – 4X – 4 + 6Y + 6 + 3 = 0

∴ X^{2} + 2X + 1 + Y^{2} + 2Y + 1 – 4X – 4 + 6Y + 6 + 3 = 0

∴ X^{2} + Y^{2} – 2X + 8Y + 7 = 0

**Ans:** The new equation of locus is X^{2} + Y^{2} – 2X + 8Y + 7 = 0

#### Example – 05:

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

The old equation of locus is 2x^{2} + 3xy – 9y^{2} – 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(X^{2} + 4X + 4) + 3(XY – X + 2Y – 2) – 9(Y^{2} – 2Y + 1) – 5X – 10 – 24Y + 24 – 7 = 0

∴ 2X^{2} + 8X + 8 + 3XY – 3X + 6Y – 6 – 9Y^{2} + 18Y – 9 – 5X – 10 – 24Y + 24 – 7 = 0

∴ 2X^{2} + 3XY – 9Y^{2} = 0

**Ans:** The new equation of locus is 2X^{2} + 3XY – 9Y^{2} = 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}= r^{2}axes remaining parallel. **Solution:**

The old equation of locus is (x – a)^{2} + (y – b)^{2} = r^{2} …………. (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} = r^{2}

∴ (X – c)^{2} + (Y)^{2} = r^{2}

∴ (X – c)^{2} + (Y)^{2} = r^{2}

∴ X^{2} – 2cX + c^{2} + Y^{2} = r^{2}

∴ X^{2} + Y^{2} – 2cX + c^{2} – r^{2} = 0

**Ans:** The new equation of locus is X^{2} + Y^{2} – 2cX + c^{2} – r^{2} = 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

∴ 4X^{2} + Y^{2} = 4

**Ans:** The new equation of locus is 4X^{2} + Y^{2} = 4

#### Example – 08:

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

The old equation of locus is 4x^{2} + y^{2} + 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(X^{2} – 2X + 1) + (Y^{2} + 4Y + 4) + 8X – 8 – 4Y – 8 + 4 = 0

∴ 4X^{2} – 8X + 4 + Y^{2} + 4Y + 4 + 8X – 8 – 4Y – 8 + 4 = 0

∴ 4X^{2} + Y^{2} – 4 = 0

∴ 4X^{2} + Y^{2} = 4

**Ans:** The new equation of locus is 4X^{2} + Y^{2} = 4

#### Example – 09:

- Obtain the new equation of locus (a – b)(x
^{2}+ y^{2}) – 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)(x^{2} + y^{2}) – 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}(X^{2} + Y^{2}) – a^{2}b^{2 }= 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 Y
^{2}= 6X. Find the equation of locus in the old system **Solution:**

The new equation of locus is Y^{2} = 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)

∴ y^{2} + 6y + 9 = 6x – 30

∴ y^{2} – 6x + 6y + 39 = 0

**Ans:** The equation of locus in old system is y^{2} – 6x + 6y + 39 = 0

#### Example – 11:

- The origin is shifted to the point (-1, 2). The equation of locus in new system is X
^{2}+ 5XY + 3Y^{2}= 0. Find the equation of locus in the old system **Solution:**

The new equation of locus is X^{2} + 5XY + 3Y^{2} = 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

∴ (x^{2} + 2x + 1) + 5(xy – 2x + y – 2) + 3(y^{2} – 4y + 4) = 0

∴ x^{2} + 2x + 1 + 5xy – 10x + 5y – 10 + 3y^{2} – 12y + 12 = 0

∴ x^{2} + 5xy + 3y^{2} – 8x – 7y + 3 = 0

**Ans:** The equation of locus in old system is x^{2} + 5xy + 3y^{2} – 8x – 7y + 3 = 0

Science > Mathematics > Locus > You are Here |

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Chemistry |
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