Physics |
Chemistry |
Biology |
Mathematics |

Science > Mathematics > Locus > You are Here |

### To Find Point at Which Origin is Shifted When Old and New Equations of Locus Are Given

#### Example – 01:

- If the equation x
^{2}– 4x – 6y + 10 = 0 is transformed to X^{2}+ AY = 0 after shifting the origin and axes remaining parallel. Find the coordinates of the point where the origin is shifted and the value of A. **Solution:**

The old equation of locus is x^{2} – 4x – 6y + 10 = 0 …………. (1)

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

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

Substituting these values in equation (1) we have

(X + h)^{2} – 4(X + h) – 6(Y + k) + 10 = 0

∴ X^{2} + 2hX + h^{2} – 4X – 4h – 6Y – 6k + 10 = 0

∴ X^{2} + (2h – 4)X – 6Y + h^{2} – 4h – 6k + 10 = 0 …………. (2)

The new equation of locus is X^{2} + AY = 0

Thus term of X and constant term are absent

∴ 2h – 4 = 0 i.e. h = 2

Similarly

h^{2} – 4h – 6k + 10 = 0

∴ (2)^{2} – 4(2) – 6k + 10 = 0

∴ 4 – 8 – 6k + 10 = 0

∴ – 6k = – 6

∴ K = 1

Thus the origin is shifted to (2, 1)

Thus the equation (2) reduces to

X^{2} – 6Y = 0

Comparing with given new equation of locus, A = – 6

**Ans:** origin is shifted to (2,1) and A = -6

#### Example – 02:

- If the equation xy – 3x + 2y – 7 = 0 is transformed to XY = 1 after shifting the origin and axes remaining parallel. Find the coordinates
**of the point where the origin is shifted.** **Solution:**

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

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

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

Substituting these values in equation (1) we have

(X + h) (Y + k) – 3(X + h) + 2(Y + k) – 7 = 0

∴ XY + kX + hY + hk – 3X – 3h + 2Y + 2k – 7 = 0

∴ XY + (k – 3)X + (h + 2)Y + (hk– 3h + 2k – 7) = 0 …………. (2)

The new equation of locus is XY = 1

Thus terms of X and Y are absent

∴ K – 3 = 0 and h + 2 = 0 and (hk– 3h + 2k – 7) = -1

∴ K = 3 and h = -2

The value of (hk– 3h + 2k – 7) = (-2)(3) – 3(-2) + 2(3) – 7 = -6 + 6 + 6 – 7 = -1

Thus values of h and k satisfy the third relation.

**Ans:** The origin is shifted to (-2, 3)

#### Example – 03:

- By shifting the origin to a suitable point O’(h, k) axes remaining parallel. The equation 2x
^{2}+ 2y^{2}-2x + 2y – 1 = 0 reduces to form X^{2}+ Y^{2}= a^{2 }(a > 0). Find O’(h, k) and value of a. **Solution:**

The old equation of locus is 2x^{2} + 2y^{2} -2x + 2y – 1 = 0…………. (1)

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

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

Substituting these values in equation (1) we have

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

∴ 2(X^{2}+ 2hX + h^{2}) + 2(Y^{2} + 2kY + k^{2}) – 2X – 2h + 2Y + 2k – 1 = 0

∴ 2X^{2}+ 4hX + 2h^{2} + 2Y^{2} + 4kY + 2k^{2} – 2X – 2h + 2Y + 2k – 1 = 0

∴ 2X^{2} + 2Y^{2} + (4h – 2)X + (4k + 2)Y + 2h^{2} + 2k^{2} – 2h + 2k – 1 = 0

∴ X^{2} + Y^{2} + (2h – 1)X + (2k + 1)Y + h^{2} + k^{2} – h + k – 1/2 = 0 …………. (2)

The new equation of locus is X^{2} + Y^{2} = a^{2}

Thus terms of X and Y are absent

∴ 2h – 1 = 0 and 2k + 1 = 0 and + h^{2} + k^{2} – h + k – 1/2 = – a^{2}

∴ h = 1/2 and h = -1/2

The origin is shifted to (1/2, -1/2)

The value of h^{2} + k^{2} – h + k – 1/2 = – a^{2}

∴ (1/2)^{2} + (-1/2)^{2} – (1/2) + (-1/2) – 1/2 = – a^{2}

∴ 1/4 + 1/4 – 1/2 – 1/2 – 1/2 = – a^{2}

∴ – 1/2 – 1/2 = – a^{2}

∴ – 1 = – a^{2}

∴ a^{2 }= 1

∴ a = + – 1

∴ a = 1 (since a> 0 given)

**Ans:** The origin is shifted to O’(1/2, -1/2) and a = 1

#### Example – 04:

- By shifting the origin to a suitable point axes remaining parallel. The equation 5y
^{2}– 9x^{2}+ 30 y + 18 x – 9 = 0 reduces to form Y^{2}/ b^{2 }– X^{2}/ a^{2 }= 1 (a > 0 and b > 0). Find the coordinates of point at which the origin is shifted. Also find values of a and b. **Solution:**

The old equation of locus is 5y^{2} – 9x^{2} + 30 y + 18 x – 9 = 0…………. (1)

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

Substituting these values in equation (1) we have

5(Y + k)^{2} – 9(X + h)^{2} + 3 (Y + k) + 18(X + h) – 9 = 0

∴ 5(Y^{2} + 2kY + k^{2}) – 9(X^{2}+ 2hX + h^{2}) + 30Y + 30k + 18X + 18h – 9 = 0

∴ 5Y^{2} + 10kY + 5k^{2} – 9X^{2 }– 18hX – 9h^{2} + 30Y + 30k + 18X + 18h – 9 = 0

∴ 5Y^{2} – 9X^{2 }+ (10k + 30)Y + (- 18h + 18)X + (5k^{2} – 9h^{2} + 30k + 18h – 9) = 0 …………. (2)

The new equation of locus is Y^{2}/ b^{2 }+ X^{2}/ a^{2 }= 1

Thus terms of X and Y are absent

∴ 10k + 30 = 0 and – 18h + 18 = 0

∴ K = -3 and h = 1

∴ The origin is shifted to (1, -3)

The value of 5k^{2} – 9h^{2} + 30k + 18h – 9 = 5(-3)^{ 2} -9(1)^{ 2} + 30(-3) + 18(1) – 9

∴ 5k^{2} – 9h^{2} + 30k + 18h – 9 = 45 – 9 – 90 + 18 – 9 = – 45

Equation (2) reduces to

5Y^{2} – 9X^{2 }– 45 = 0

∴ 5Y^{2} – 9X^{2 }= 45

∴ 5Y^{2}/45 – 9X^{2}/45 = 1

∴ Y^{2}/9 – X^{2}/5 = 1

Comparing with Y^{2}/ b^{2 }– X^{2}/ a^{2 }= 1

∴ b^{2 }= 9 and a^{2 }= 5

Hence b = 3 and a = √5 (only positive values are considered since a> o and b > 0)

**Ans:** The origin is shifted to (1, -3) and a = √5 and b = 3

#### Example – 05:

- By shifting the origin to O'(h, k) axes remaining parallel, reduce the equation 4x
^{2}+ 9y^{2}+16x – 18 y +24 = 0 reduces to form X^{2}/ a^{2 }+ Y^{2}/ b^{2 }= 1 (a > 0 and b > 0). Find O'(h, k) at which the origin is shifted. Also find values of a and b. **Solution:**

The old equation of locus is 4x^{2} + 9y^{2} +16x – 18 y +24 = 0 …………. (1)

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

Substituting these values in equation (1) we have

4(X + h)^{2} + 9(Y + k)^{2} +16(X + h) – 18(Y + k) +24 = 0

∴ 4(X^{2}+ 2hX + h^{2}) + 9(Y^{2} + 2kY + k^{2}) +16(X + h) – 18(Y + k) +24 = 0

∴ 4X^{2}+ 8hX + 4h^{2} + 9Y^{2} + 18kY + 9k^{2} +16X + 16h – 18Y – 18k +24 = 0

∴ 4X^{2 }+ 9Y^{2} + (8h + 16)X + (18k – 18)Y + (4h^{2 }+ 9k^{2} + 16h – 18k +24) = 0 …………. (2)

The new equation of locus is X^{2}/ a^{2 }+ Y^{2}/ b^{2 }= 1

Thus terms of X and Y are absent

∴ 8h + 16 = 0 and 18k – 18 = 0

∴ h = -2 and k = 1

∴ The origin is shifted to O'(-2, 1)

The value of (4h^{2 }+ 9k^{2} + 16h – 18k +24) = 4(-2)^{2 }+ 9(1)^{2} + 16(-2) – 18(1) +24

∴ 4h^{2 }+ 9k^{2} + 16h – 18k +24 = 16 + 9 -32 – 18 + 24 = – 1

Equation (2) reduces to

4X^{2}+ 9Y^{2 }– 1 = 0

∴ 4X^{2}+ 9Y^{2 }= 1

∴ X^{2}/(1/4) + Y^{2}/(1/9) = 1

Comparing with X^{2}/ a^{2 }+ Y^{2}/ b^{2 }= 1

∴ a^{2 }= 1/4 and b^{2 }= 1/9

Hence a = 1/2 and b = 1/3 (only positive values are considered since a> o and b > 0)

**Ans:** The origin is shifted to O'(-2, 1) and a = 1/2 and b = 1/3

#### Example – 06:

- By shifting the origin to suitable point axes remaining parallel, reduce the equation 2x
^{2}– y^{2}– 4x + 4y – 3 = 0 reduces to form X^{2}/ a^{2 }– Y^{2}/ b^{2 }= 1 (a > 0 and b > 0). Find the coordinates of point at which the origin is shifted. Also find values of a and b. **Solution:**

The old equation of locus is 2x^{2} – y^{2} – 4x + 4y – 3 = 0 …………. (1)

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

Substituting these values in equation (1) we have

2(X + h)^{2} – (Y + k)^{2} – 4(X + h) + 4(Y + k) – 3 = 0

∴ 2(X^{2}+ 2hX + h^{2}) – (Y^{2} + 2kY + k^{2}) – 4(X + h) + 4(Y + k) – 3 = 0

∴ 2X^{2}+ 4hX + 2h^{2} – Y^{2} – 2kY – k^{2} – 4X – 4h + 4Y + 4k – 3 = 0

∴ 2X^{2 }– Y^{2} + (4h – 4)X + (-2k +4)Y + (2h^{2} – k^{2} – 4h + 4k – 3) = 0…………. (2)

The new equation of locus is X^{2}/ a^{2 }– Y^{2}/ b^{2 }= 1

Thus terms of X and Y are absent

∴ 4h – 4 = 0 and -2k +4 = 0

∴ h = 1 and k = 2

∴ The origin is shifted to (1, 2)

The value of (2h^{2} – k^{2} – 4h + 4k – 3) = 2(1)^{2} – (2)^{2} – 4(1) + 4(2) – 3

∴ 2h^{2} – k^{2} – 4h + 4k – 3 = 2 – 4 – 4 + 8 – 3 = -1

Equation (2) reduces to

2X^{2}– Y^{2 }– 1 = 0

∴ 2X^{2}– Y^{2 }= 1

∴ X^{2}/(1/2) + Y^{2}/1 = 1

Comparing with X^{2}/ a^{2 }+ Y^{2}/ b^{2 }= 1

∴ a^{2 }= 1/2 and b^{2 }= 1

Hence a = 1/√2 and b = 1 (only positive values are considered since a> o and b > 0)

**Ans:** The origin is shifted to (1, 2) and a = 1/√2 and b = 1

#### Example – 07:

- By shifting the origin to suitable point axes remaining parallel, the equation 3x
^{2}– 5y^{2}– 6x – 20y – 32 = 0 reduces to form X^{2}/ a^{2 }– Y^{2}/ b^{2 }= 1 . Find the coordinates of point at which the origin is shifted. Also find values of a and b. **Solution:**

The old equation of locus is 3x^{2} – 5y^{2} – 6x – 20y – 32 = 0 …………. (1)

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

Substituting these values in equation (1) we have

3(X + h)^{2} – 5(Y + k)^{2} – 6(X + h) – 20(Y + k) – 32 = 0

∴ 3(X^{2}+ 2hX + h^{2}) – 5(Y^{2} + 2kY + k^{2}) – 6(X + h) – 20(Y + k) -32 = 0

∴ 3X^{2}+ 6hX + 3h^{2} – 5Y^{2} – 10kY – 5k^{2} – 6X – 6h – 20Y – 20k -32 = 0

∴ 3X^{2 }– 5Y^{2} + (6h – 6)X + (- 10k – 20) Y + (3h^{2} – 5k^{2} – 6h – 20k -32) = 0…………. (2)

The new equation of locus is X^{2}/ a^{2 }– Y^{2}/ b^{2 }= 1

Thus terms of X and Y are absent

∴ 6h – 6 = 0 and – 10k – 20 = 0

∴ h = 1 and k = -2

∴ The origin is shifted to (1, -2)

The value of (3h^{2} – 5k^{2} – 6h – 20k -32) =3(1)^{2} – 5(-2)^{2} – 6(1) – 20(-2) -32

∴ 3h^{2} – 5k^{2} – 6h – 20k -32 = 3 – 20 – 6 + 40 -32 = -15

Equation (2) reduces to

3X^{2 }– 5Y^{2 }– 15 = 0

∴ 3X^{2 }– 5Y^{2 }= 15

∴ 3X^{2}/15 – 5Y^{2}/15 = 1

∴ X^{2}/5 – Y^{2}/3 = 1

Comparing with X^{2}/ a^{2 }+ Y^{2}/ b^{2 }= 1

∴ a^{2 }= 5 and a^{2 }= 3

Hence a = √5 and b = √3

**Ans:** The origin is shifted to (1, -2) and a = √5 and b = √3

#### Example – 08:

- By shifting the origin to suitable point axes remaining parallel, the equation y
^{2}– 6x + 4y + 28 = 0 reduces to form Y^{2}^{ }= 4aX . Find the coordinates of point at which the origin is shifted. Also find value of a **Solution:**

The old equation of locus is y^{2} – 6x + 4y + 28 = 0 …………. (1)

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

Substituting these values in equation (1) we have

(Y + k)^{2} – 6(X + h) + 4(Y + k) + 28 = 0

∴ Y^{2} + 2kY + k^{2} – 6X – 6h + 4Y + 4k + 28 = 0

∴ Y^{2} + (2k + 4)Y + (k^{2} – 6h + 4k + 28) = 6X …………. (2)

The new equation of locus is Y^{2} = 4ax

Thus terms of Y and constant term are absent

∴ 2k + 4 = 0 and k^{2} – 6h + 4k + 28 = 0

∴ k = -2

and (-2)^{2} – 6h + 4(-2) + 28 = 0

∴ 4 – 6h – 8 + 28 = 0

∴ – 6h = – 24

∴ h = 4

∴ The origin is shifted to (4, -2)

Equation (2) reduces to

Y^{2 }= 6x

Comparing with Y^{2 }= 4ax

∴ 4a = 6 i.e. a = 3/2

**Ans:** The origin is shifted to (4, -2) and a = 3/2

#### Example – 09:

- By shifting the origin to suitable point axes remaining parallel, reduce the equation y
^{2}+ 8x + 4y – 2 = 0 such that it will not contain term in y and constant term . Find the coordinates of point at which the origin is shifted. **Solution:**

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

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

Substituting these values in equation (1) we have

(Y + k)^{2} + 8(X + h) + 4(Y + k) – 2 = 0

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

∴ Y^{2} + (2k + 4)Y + (k^{2} + 8h + 4k – 2) + 8X = 0 …………. (2)

Thus terms of Y and constant term are absent

∴ 2k + 4 = 0 and k^{2} + 8h + 4k -2 = 0

∴ k = -2

and (-2)^{2} + 8h + 4(-2) – 2 = 0

∴ 4 + 8h – 8 – 2 = 0

∴ 8h = 6

∴ h = 3/4

∴ The origin is shifted to (3/4, -2)

**Ans:** The origin is shifted to (3/4, -2)

#### Example – 10:

- The origin is shifted to point (3, p). Find value of p so that the new equation of locus y
^{2}+ 2x – 8y + 7 = 0 such that it will not contain term in y. Find value of p. **Solution:**

The old equation of locus is y^{2} + 2x – 8y + 7 = 0 …………. (1)

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

We have x = X + 3 and y = Y + p

Substituting these values in equation (1) we have

y^{2} + 2x – 8y + 7 = 0

(Y + p)^{2} + 2(X + 3) – 8(Y + p) + 7 = 0

∴ Y^{2} + 2pY + p^{2} + 2X + 6 – 8Y – 8p + 7 = 0

∴ Y^{2} + (2p – 8)Y + (p^{2} – 8p + 13) + 2X = 0 …………. (2)

Thus terms of Y and constant term are absent

∴ 2p – 8 = 0

∴ p = 4

Science > Mathematics > Locus > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |