Question

Locus of a point 'P' such that PA + PB = 4,
where A = (2, 3, 4), B = (-2, 3, 4) is
(A) y2 – 22 +6y - 8z + 25 = 0
(B) y2 + x2 – 6y - 82 + 25 = 0
(C) y? – x2 + 6y – 8z – 25 = 0
(D) y2 + 22 – 6y + 8z + 25 = 0​

Answer 1

PA + PB = 4

Let ,The Co-ordinates Of P are P = ( x,y,z )

The Formula To Finding The Distance In 3d Space is , Distance = √[(x2-x1)²+(y2-y1)²+(z2-z1)²]

As we Have Given , A = (2, 3, 4), B = (-2, 3, 4)

Locus is The set of Points satisfies The Equation

Therefore, PA+PB=4

↠ √[(2-x)²+(3-y)²+(4-z)²]+√[(-2-x)²+(3-y)²+(4- z)²]=4

↠ √[(2-x)²+(3-y)²+(4-z)²] = 4 - √[(-2-x)²+(3-y)²+(4-z)²]

Squaring On Both The Sides :

↠ (2-x)²+(3-y)²+(4-z)²=16-8√[(-2-x)²+(3-y)²+(4-z)²]+(-2-x)²+(3-y)²+(4-z)²

↠ (2-x)²=16-8√[(-2-x)²+(3-y)²+(4-z)²]+(-2-x)²

↠. (2-x)²-(-2-x)²=16-8√[(-2-x)²+(3-y)²+(4-z)²]

↠ 4-4x+x²-(4+4x+x²) = 16-8√[(-2-x)²+(3-y)²+(4-z)²]

↠ 4-4x+x²-4-4x-x² = 16-8√[(-2-x)²+(3-y)²+(4-z)²]

↠ -8x=16-8√[(2-x)²+(3-y)²+(4-z)²]

↠ -8(x+2)=-8√[(2-x)²+(3-y)²+(4-z)²]

↠ (x+2) = √[(2-x)²+(3-y)²+(4-z)²]

Again Squaring on Both The Sides :

↠ (x+2)² = (-2-x)²+(3-y)²+(4-z)²

↠ x²+4x+4 = 4 + 4x + x² + 9 - 6y + y² + 16 - 8z + z²

↠ 9- 6y +y² + 16 - 8z + z² = 0

y² + z² - 6y - 8z + 25 = 0