To Find Point at Which Origin is Shifted

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

Example – 01:

  • If the equation x2 – 4x – 6y + 10 = 0 is transformed to X2 + 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 x2 – 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

∴   X2 + 2hX + h2 – 4X – 4h – 6Y – 6k + 10 = 0

∴   X2 + (2h – 4)X – 6Y  + h2 – 4h – 6k + 10 = 0 …………. (2)

The new equation of locus is X2 + AY = 0



Thus term of X and constant term are absent

∴   2h – 4 = 0 i.e. h = 2

Similarly

h2 – 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



X2 – 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 2x2 + 2y2 -2x + 2y – 1 = 0 reduces to form X2 + Y2 = a(a > 0). Find O’(h, k) and value of a.
  • Solution:

The old equation of locus is 2x2 + 2y2 -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(X2+ 2hX + h2) + 2(Y2 + 2kY + k2) – 2X – 2h + 2Y + 2k – 1 = 0

∴   2X2+ 4hX + 2h2 + 2Y2 + 4kY + 2k2 – 2X – 2h + 2Y + 2k – 1 = 0

∴   2X2 + 2Y2 + (4h – 2)X + (4k + 2)Y + 2h2 + 2k2 – 2h + 2k – 1 = 0



∴   X2 + Y2 + (2h – 1)X + (2k + 1)Y + h2 + k2 – h + k – 1/2 = 0  …………. (2)

The new equation of locus is X2 + Y2 = a2

Thus terms of X and Y are absent

∴   2h – 1 = 0 and 2k + 1 = 0 and + h2 + k2 – h + k – 1/2 = – a2

∴   h = 1/2 and h = -1/2

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

The value of h2 + k2 – h + k – 1/2 = – a2

∴   (1/2)2 + (-1/2)2 – (1/2) + (-1/2) – 1/2 = – a2

∴   1/4 + 1/4 – 1/2 – 1/2 – 1/2 = – a2

∴   – 1/2 – 1/2 = – a2

∴   – 1 = – a2

∴   a2 = 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 5y2 – 9x2 + 30 y + 18 x – 9 = 0 reduces to form Y2/ b2 – X2/ a2 = 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 5y2 – 9x2 + 30 y + 18 x – 9 = 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

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

∴   5(Y2 + 2kY + k2) – 9(X2+ 2hX + h2) + 30Y + 30k + 18X + 18h – 9 = 0

∴   5Y2 + 10kY + 5k2 – 9X2 – 18hX – 9h2 + 30Y + 30k + 18X + 18h – 9 = 0

∴   5Y2 – 9X2 + (10k + 30)Y + (- 18h + 18)X + (5k2 – 9h2 + 30k + 18h – 9) = 0 …………. (2)

The new equation of locus is Y2/ b2 + X2/ a2 = 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 5k2 – 9h2 + 30k + 18h – 9 = 5(-3) 2 -9(1) 2 + 30(-3) + 18(1) – 9

∴   5k2 – 9h2 + 30k + 18h – 9  =  45 – 9 – 90 + 18  – 9 = – 45

Equation (2) reduces to

5Y2 – 9X2 – 45 = 0

∴   5Y2 – 9X2 = 45

∴   5Y2/45 – 9X2/45 = 1

∴   Y2/9 – X2/5 = 1

Comparing with Y2/ b2 – X2/ a2 = 1

∴   b2 = 9 and a2 = 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 4x2 + 9y2 +16x – 18 y +24 = 0 reduces to form X2/ a2 + Y2/ b2 = 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 4x2 + 9y2 +16x – 18 y +24 = 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

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

∴  4(X2+ 2hX + h2) + 9(Y2 + 2kY + k2) +16(X + h) – 18(Y + k) +24 = 0

∴  4X2+ 8hX + 4h2 + 9Y2 + 18kY + 9k2 +16X + 16h – 18Y – 18k +24 = 0

∴  4X+ 9Y2  + (8h + 16)X  + (18k – 18)Y + (4h+ 9k2 + 16h – 18k +24) = 0 …………. (2)

The new equation of locus is X2/ a2 + Y2/ b2 = 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+ 9k2 + 16h – 18k +24) = 4(-2)+ 9(1)2 + 16(-2) – 18(1) +24

∴  4h+ 9k2 + 16h – 18k +24  =  16 + 9 -32 – 18 + 24 = – 1

Equation (2) reduces to

4X2+ 9Y2 – 1 = 0

∴  4X2+ 9Y2 = 1

∴  X2/(1/4) + Y2/(1/9) = 1

Comparing with X2/ a2 + Y2/ b2 = 1

∴   a2 = 1/4 and b2 = 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 2x2 – y2 – 4x + 4y – 3 = 0 reduces to form X2/ a2 – Y2/ b2 = 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 2x2 – y2 – 4x + 4y – 3 = 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 – (Y + k)2 – 4(X + h) + 4(Y + k) – 3 = 0

∴  2(X2+ 2hX + h2) – (Y2 + 2kY + k2) – 4(X + h) + 4(Y + k) – 3 = 0

∴  2X2+ 4hX + 2h2 – Y2 – 2kY – k2 – 4X – 4h + 4Y + 4k – 3 = 0

∴  2X– Y2 + (4h – 4)X + (-2k +4)Y + (2h2 – k2 – 4h + 4k – 3) = 0…………. (2)

The new equation of locus is X2/ a2 – Y2/ b2 = 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 (2h2 – k2 – 4h + 4k – 3) = 2(1)2 – (2)2 – 4(1) + 4(2) – 3

∴  2h2 – k2 – 4h + 4k – 3  =  2 – 4 – 4 + 8 – 3 = -1

Equation (2) reduces to

2X2– Y2 – 1 = 0

∴  2X2– Y2 = 1

∴  X2/(1/2) + Y2/1 = 1

Comparing with X2/ a2 + Y2/ b2 = 1

∴   a2 = 1/2 and b2 = 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 3x2 – 5y2 –  6x – 20y – 32 = 0 reduces to form X2/ a2 – Y2/ b2 = 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 3x2 – 5y2 –  6x – 20y – 32 = 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

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

∴  3(X2+ 2hX + h2) – 5(Y2 + 2kY + k2) – 6(X + h) – 20(Y + k) -32 = 0

∴  3X2+ 6hX + 3h2 – 5Y2 – 10kY – 5k2 – 6X – 6h – 20Y – 20k -32 = 0

∴   3X– 5Y2 + (6h – 6)X + (- 10k – 20) Y + (3h2  – 5k2 – 6h – 20k -32) = 0…………. (2)

The new equation of locus is X2/ a2 – Y2/ b2 = 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 (3h2  – 5k2 – 6h – 20k -32)  =3(1)2  – 5(-2)2 – 6(1) – 20(-2) -32

∴  3h2  – 5k2 – 6h – 20k -32  =  3  – 20 – 6 + 40 -32 = -15

Equation (2) reduces to

3X– 5Y2 – 15 = 0

∴  3X– 5Y2 = 15

∴  3X2/15 – 5Y2/15 = 1

∴  X2/5 – Y2/3 = 1

Comparing with X2/ a2 + Y2/ b2 = 1

∴   a2 = 5 and a2 = 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 y2 –  6x + 4y + 28 = 0 reduces to form Y2 = 4aX . Find the coordinates of point at which the origin is shifted. Also find value of a
  • Solution:

The old equation of locus is y2 –  6x + 4y + 28 = 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

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

∴  Y2 + 2kY + k2 –  6X – 6h + 4Y + 4k + 28 = 0

∴  Y2 + (2k + 4)Y + (k2 – 6h + 4k + 28) = 6X …………. (2)

The new equation of locus is Y2 = 4ax

Thus terms of Y and constant term are absent

∴  2k + 4 = 0 and k2 – 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

Y2  = 6x

Comparing with Y= 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 y2 +  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 y2 +  8x + 4y – 2 = 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

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

∴  Y2 + 2kY + k2 +  8X + 8h + 4Y + 4k – 2 = 0

∴  Y2 + (2k + 4)Y + (k2 + 8h + 4k – 2) + 8X = 0 …………. (2)

Thus terms of Y and constant term are absent

∴  2k + 4 = 0 and k2 + 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 y2 +  2x – 8y + 7 = 0 such that it will not contain term in y. Find value of p.
  • Solution:

The old equation of locus is y2 +  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

y2 +  2x – 8y + 7 = 0

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

∴  Y2 + 2pY + p2 +  2X + 6 – 8Y – 8p + 7 = 0

∴  Y2 + (2p – 8)Y + (p2 – 8p + 13) + 2X = 0 …………. (2)

Thus terms of Y and constant term are absent

∴  2p – 8 = 0

∴   p = 4

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