# Equation of Locus 02

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#### Example – 31 :

• Find the equation of the locus of a point such that its distance from (5, 0) is equal to a distance from the y-axis.
• Solution:

Let P(x. y) be the point on the locus, A(5. 0) be the point

Distance of point from y-axis = x

Given: AP = x

∴ AP² = x²

∴ [(x – 5 )²  + (y – 0)² ] = x²

∴ x ²  – 10x + 25 + y ²  = x²

∴ – 10x + 25 + y ²  =0

∴   y ² – 10x + 25   =0

Hence required equation of the locus  is  y ² – 10x + 25   =0

#### Example – 32 :

• Find the equation of the locus of a point which is equidistant from (1, 3) and the x-axis.
• Solution:

Let P(x. y) be the point on the locus, A(1, 3) be the point

Distance of point from x-axis = y

Given: AP = y

∴ AP² = y²

∴ [(x – 1 )²  + (y – 3)² ] = y²

∴ x ²  – 2x + 1 + y ² – 6y + 9  = y²

∴ x ²  – 2x  – 6y + 10  = 0

Hence required equation of the locus  is x ²  – 2x  – 6y + 10  = 0

#### Example – 33:

• Find the equation of locus of a point such that, the sum of square of its distances from the points (2, 5) and (3, -1) is 40.
• Solution:

Let P(x. y) be the point on the locus, Given A(2, 5) and B(3, -1) be the given points

Given  PA² + PB² = 40

∴ [(x – 2 )²  + (y – 5)² ] +  [(x – 3)²  + (y + 1)²] = 40

∴ x²  – 4x + 4 + y² – 10y + 25 + x² – 6x + 9  + y² + 2y + 1 = 40

∴ 2x² + 2y²  – 10x   – 8y – 1 = 0

Hence required equation of the locus  is 2x² + 2y²  – 10x   – 8y – 1 = 0

#### Example – 34:

• Find the equation of locus of a point such that, its distance from (a, 0) is m times  its distance from (0, a)
• Solution:

Let P(x. y) be the point on the locus, Given A(a, o) and B(0, a) be the given points

Given  PA = m. PB

∴   PA² = m². PB²

∴ [(x – a )²  + (y – 0)² ] =  m²[(x – 0)²  + (y – a)²]

∴ x²  – 2ax + a² + y² = m²( x² + y² – 2ay + a²)

∴ x²  – 2ax + a² + y² = m²x² + m²y² – 2am²y + a²m²

∴ x²  – 2ax + a² + y² – m²x² – m²y² + 2am²y – a²m² = 0

∴ (1 – m²) x²  + ( 1 – m²) y²  – 2ax  + 2am²y +  (1 – m²) a² = 0

∴ (1 – m²)( x²  +  y² ) – 2ax  + 2am²y +  a²(1 – m²) = 0

Hence required equation of the locus  is (1 – m²)( x²  +  y² ) – 2ax  + 2am²y +  a²(1 – m²) = 0

#### Example – 35:

• The point S is (3, 0) and abscissa of point M is -3. A variable point P is such that ordinate of P and M are equal. Find the equation of locus of point P such that SP = PM.
• Solution:

Let P(x. y) be the point on the locus, Given S(3,0)

The abscissa of point M is -3 and its ordinate is same as P i.e. y. Hence M(-3, y) is the point

Given  SP = PM

∴  SP² = PM²

∴ [(x – 3 )²  + (y – 0)² ] =  [(x + 3)²  + (y – y)²]

∴ x²  – 6x + 9 + y² =  x² + 6x + 9  + 0

∴ x²  – 6x + 9 + y² –  x² – 6x – 9  = 0

∴ – 12x   + y² = 0

∴  y² – 12x = 0

Hence required equation of the locus  is y² – 12x = 0

#### Example – 36:

• Find the equation of locus of a point which moves such that the ratio of its distances from (2, 0) and (1, 3) is 5:4.
• Solution:

Let P(x. y) be the point on the locus and A(2,0) and B(1, 3) be the points

Given  PA/ PB = 5/4

∴  4 PA = 5 PB

∴  16 PA² = 25 PB²

∴ 16 [(x – 2)²  + (y – 0)² ] =  25[(x – 1)²  + (y – 3)²]

∴ 16(x²  – 4x + 4 + y² ) =  25(x² – 2x + 1  + y² – 6y + 9)

∴ 16x²  – 64x + 64 + 16y²  =  25x² – 50x + 25  + 25 y² – 150y + 225

∴  25x² – 50x + 25  + 25 y² – 150y + 225 – 16x²  + 64x – 64 – 16y²  = 0

∴  9x² + 9 y²  + 14x  – 150y – 186  = 0

Hence required equation of the locus  is 9x² + 9 y²  + 14x  – 150y – 186  = 0

#### Example – 37:

• Find the equation of locus of a point such that the sum of its distances from co-orinate axes is thrice its distance from the origin.
• Solution:

Let P(x. y) be the point on the locus and O(0,0) be the origin

Distance of point from y-axis = x

Distance of point from x-axis = y

Given  x + y = 3OP

(x + y)² = 9OP²

∴ x² + 2xy + y²=  9[(x – 0)²  + (y – 0)²]

∴ x² + 2xy + y² =  9x ²  + 9y²

∴   9x ²  + 9y² – x² – 2xy – y² = 0

∴   8x ²  + 8y² – 2xy  = 0

∴   4x ²  + 4y² – xy  = 0

Hence required equation of the locus  is 4x ² – xy + 4y²= 0

#### Example – 38:

• A(4, 0) and B(-4, 0) are two given points. A variable point P such that PA + PB = 10, show that the equation of locus of point P is • Solution:

Let P(x. y) be the point on the locus and A(4, 0) and B(-4, 0) be the points

Given  PA  + PB = 10

∴  √(x – 4)²  + ( y – 0)²   + (x + 4)²  + ( y – 0)²   =10

∴  √x² – 8x + 16  + y ²   + x² + 8x + 16  + y²   =10

∴  √x² – 8x + 16  + y ²   = 10 – x² + 8x + 16  + y²

Squaring both sides

∴  x² – 8x + 16  + y ² = 100 – 20 x² + 8x + 16  + y²  + x² + 8x + 16  + y ²

∴  – 8x  = 100 – 20 x² + 8x + 16  + y²   + 8x

∴  20 x² + 8x + 16  + y²   =100 + 16x

Squaring both sides

400(x² + 8x + 16  + y ²) = 10000 + 3200x + 256x²

∴   400x² + 3200x + 6400  + 400y ² = 10000 + 3200x + 256x²

∴   400x² – 256x² + 400y ² = 10000  – 6400

∴   144x² + 400y ² = 3600

Dividing both sides by 3600 Proved as required

#### Example – 39:

• A(5, 0) and B(-5, 0) are two given points. A variable point P such that PA – PB = 6, show that the equation of locus of point P is • Solution:

Let P(x. y) be the point on the locus and A(5, 0) and B(-5, 0) be the points

Given  PA  –  PB = 6

∴  √(x – 5)²  + ( y – 0)²  (x + 5)²  + ( y – 0)²   =6

∴  √x² – 10x + 25  + y ²  x² + 10x + 25  + y²   = 6

∴  √x² – 10x + 25  + y ²   = 6 + x² + 10x + 25  + y²

Squaring both sides

∴  x² – 10x + 25  + y ² = 36 + 12 x² + 10x + 25  + y²  + x² + 10x + 25  + y ²

∴  – 10x  = 36 + 12 x² + 10x + 25  + y²   + 10x

∴  – 12 x² + 10x + 25  + y²   = 36 + 20x

Squaring both sides

144(x² + 10x + 25 + y ²) = 1296 + 1440x + 400x²

∴  144x² + 1440x + 3600 + 144 y ² = 1296 + 1440x + 400x²

∴   400x² – 144x² – 144 y ² = 3600 – 1296

∴   256x² – 144 y ² = 2304

Dividing both sides by 2304 Proved as required

#### Example – 40:

• A(-3, 0) and B(3, 0) are two given points. A variable point P such that AP – PB = 10, show that the equation of locus of point P is • Solution:

Let P(x. y) be the point on the locus and A(-3, 0) and B(3, 0) be the points

Given  AP – PB = 10

∴  √(x + 3)²  + ( y – 0)²  (x – 3)²  + ( y – 0)²   =10

∴  √x² + 6x + 9  + y ²  x² – 6x + 9  + y²   =10

∴  √x² +  6x + 9  + y ²   = 10 – x² – 6x + 9  + y²

Squaring both sides

∴  x² + 6x + 9  + y ² = 100 – 20 x² – 6x + 9  + y²  + x² – 6x + 9  + y ²

∴  6x  = 100 – 20 x² – 6x + 9  + y²   – 6x

∴  20 x² – 6x + 9  + y²   =100 – 12x

Squaring both sides

400(x² – 6x + 9  + y ²) = 10000 – 2400x + 144x²

∴  400x² – 2400x + 3600  + 400y ² = 10000 – 2400x + 144x²

∴   400x² – 144x² + 400y ² = 10000  – 3600

∴   256x² + 400y ² = 6400

Dividing both sides by 6400 Proved as required

#### Example – 41:

• A(2, 0) and B(-2, 0) are two given points. A variable point P such that sum of its distances from given points is 6. Find the equation of the locus of point P.
• Solution:

Let P(x. y) be the point on the locus and A(2, 0) and B(-2, 0) be the points

Given  PA  + PB = 10

∴  √(x – 2)²  + ( y – 0)²   + (x + 2)²  + ( y – 0)²   =6

∴  √x² – 4x + 4  + y ²   + x² + 4x + 4  + y²   =c6

∴  √x² – 4x + 4  + y ²   = 6 – x² + 4x + 4  + y²

Squaring both sides

∴  x² – 4x + 4  + y ² = 36 – 12 x² + 4x + 4  + y²  + x² + 4x + 4  + y ²

∴  – 4x  = 36 – 12 x² + 4x + 4  + y²   + 4x

∴  12 x² + 4x + 4  + y²   =36 + 8x

Squaring both sides

144(x² + 4x + 4  + y ²) = 1296 + 576x + 64x²

∴  144x² +  576x + 576  + 144y ² = 1296 + 576x + 64x²

∴  144x²  + 144y ² – 64x² = 1296 – 576

∴  80x²  + 144y ²= 720

Dividing both sides by 720 This is the required equation of the locus.

#### Example – 42:

• A(c, 0) and B(-c, 0) are two given points. A variable point P such that sum of its distances frpm given point is 2a. Find equation of locus if b² = a² – c².
• Solution:

Let P(x. y) be the point on the locus and A(c, 0) and B(-c, 0) be the points

Given  AP – PB = 10

∴  √(x – c)²  + ( y – 0)²  (x + c)²  + ( y – 0)²   = 2a

∴  √x² – 2cx + c²  + y ²  x²+ 2cx + c² + y²   = 2a

∴  √x² – 2cx + c²  + y ²   = 2a – x² + 2cx + c²  + y²

Squaring both sides

∴  x² – 2cx + c ²  + y ² = 4a ² – 4a x² + 2cx + c²  + y²  + x² + 2cx + c²  + y ²

∴  – 2cx  = 2a – 4a x² + 2cx + c²  + y²    + 2cx

∴  4a x² + 2cx + c²  + y²   = 4a ² + 4cx

Squaring both sides

16a²(x² + 2cx + c²  + y²) = 16a4 + 32a²cx + 16c²x²

∴  16a²x² + 32a²cx + 16a²c²  + 16a²y² = 16a4  + 32a²cx + 16c²x²

∴  16a²x² – 16c²x² + 16a²y² = 16a4  – 16a²c²

∴  16(a² – c²)x² + 16a²y² = 16a² ( a² – c²)

Given b² = a² – c²

∴  16b²x² + 16a²y² = 16a² b²

∴  b²x² + a²y² = a² b²

Dividing both sides by a² b² This is the equation of locus.

#### Example – 43 :

• Find the equation of locus of a point P so that the segment joining the points (3, 2) and (-5, 1) subtends a right angle at the point P.
• Solution:
• Method – I (Using Pythagoras Theorem):

Let P(x. y) be the point on the locus, A(3, 2) and B(-5, 1) be the points.

Seg AB subtends right angle at point P, hence ΔPAB is right-angled triangle

Seg AB is the hypotenuse

PA² +  PB² = AB²

∴  (x – 3)² + (y – 2)² + (x + 5)² + (y – 1)² = (3 + 5)² + (2 – 1)²

∴  x² – 6x + 9 + y² – 4y + 4 + x² + 10x + 25 + y² – 2y + 1 = 64 + 1

∴ 2 x² + 2y² + 4x  – 6y + 39 – 65 = 0

∴ 2 x² + 2y² + 4x  – 6y – 26 = 0

∴ x² + y² + 2x  – 3y – 13 = 0

Hence required equation of the locus  is  x² + y² + 2x  – 3y – 13 = 0

• Method – II (Using Slopes):

Let P(x. y) be the point on the locus, A(3, 2) and B(-5, 1) be the points.

Seg AB subtends right angle at point P

Slope of AP ×  Slope of BP = -1 ∴  y² – y – 2y + 2 = -(x² +5x – 3x -15)

∴  y² – 3y + 2 = – (x² + 2x -15)

∴  y² – 3y + 2 =  – x² –  2x +15

∴  y² – 3y + 2 + x² +  2x – 15 = 0

∴  x² + y² + 2x  – 3y – 13 = 0

Hence required equation of the locus  is  x² + y² + 2x  – 3y – 13 = 0

#### Example – 44 :

• A(2, 0) and B(-2, 0) are two points. Find the equation of locus of point P, such that ∠ APB is a right angle.
• Solution:
• Method – I (Using Pythagoras Theorem):

Let P(x. y) be the point on the locus, A(2, 0) and B(-2, 0) be the points.

∠ APB is a right angle, hence ΔPAB is right-angled triangle

Seg AB is the hypotenuse

PA² +  PB² = AB²

∴  (x – 2)² + (y – 0)² + (x + 2)² + (y – 0)² = (2 + 2)² + (0 – 0)²

∴  x² – 4x + 4 + y² + x² + 4x + 4 + y²  = 16

∴ 2 x² + 2y²  = 8

∴  x² +  y² = 4

Hence required equation of the locus  is    x² +  y² = 4

• Method – II (Using Slopes):

Let P(x. y) be the point on the locus, A(2, 0) and B(-2, 0) be the points.

Seg AB subtends right angle at point P

Slope of AP ×  Slope of BP = -1 ∴  y²  = -(x²  – 4)

∴  y²  = – x²  + 4

∴   x²  +  y²  = 4

Hence required equation of the locus  is   x²  +  y²  = 4

#### Example – 45 :

• A(5, -3) and B(-1, -5) are two points. Find the equation of locus of point P, such that seg AB subtends right angle at point P.
• Solution:

Let P(x. y) be the point on the locus, A(5, -3) and B(-1, -5) be the points.

The seg AB subtends right angle at point P, hence ΔPAB is right-angled triangle

Seg AB is the hypotenuse

PA² +  PB² = AB²

∴  (x – 5)² + (y + 3)² + (x + 1)² + (y + 5)² = (5 + 1)² + (-3 + 5)²

∴  x² – 10x + 25 + y² + 6y + 9 + x² + 2x + 1 + y² + 10y + 25  = 36  + 4

∴  2x² + 2y²  – 8x + 16 y + 60  = 40

∴  2x² + 2y²  – 8x + 16 y + 20 = 0

∴  x² + y²  – 4x + 8 y + 10 = 0

Hence required equation of the locus  is  x² + y²  – 4x + 8 y + 10 = 0

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