Find the locus of P if P A2-P B2-2 k2- where A and B are the points 3-4-5 and -1-3-7-

Let P (x, y, z) be the point if PA2+PB2=2k2 #8658;x 32+y 42+z 522+x+12+y 32+z+722=2k2 #8658;x2 6x+9+y2 8y+16+z2 10z+25+x2+2x+1+y2 6y+9+z2+14z+49=2k2 #86 ...

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Quantitative Aptitude

Coordinate Geometry

Find the locu...

Question

Find the locus of P if PA² + PB²= 2k², where A and B are the points (3, 4, 5) and (–1, 3, –7).

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Solution

Let P (x, y, z) be the point if $P A^{2} + P B^{2} = 2 k^{2}$

$\Rightarrow {(\sqrt{{(x - 3)}^{2} + {(y - 4)}^{2} + {(z - 5)}^{2}})}^{2} + {(\sqrt{{(x + 1)}^{2} + {(y - 3)}^{2} + {(z + 7)}^{2}})}^{2} = 2 k^{2} \Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 + z^{2} - 10 z + 25 + x^{2} + 2 x + 1 + y^{2} - 6 y + 9 + z^{2} + 14 z + 49 = 2 k^{2} \Rightarrow 2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 - 2 k^{2} = 0$

Hence, $2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 - 2 k^{2} = 0$ is the locus of point P.

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Similar questions

Q. The locus of a point P which moves such that

P A^{2} - P B^{2} = 2 k^{2}

where A and B are

(3, 4, 5)

and

(- 1, 3, - 7)

respectively is

Q. If

A

and

B

be the points

(3, 4, 5)

and

(- 1, 3, - 7)

respectively. Find the equation of the set of points

P

such that

P A^{2}

P B^{2}

= K^{2}

, where

K

is a constant

Q. The coordinates of the points

A

and

B

are

(a, 0)

and

(- a, 0)

, respectively. If a point

P

moves so that

P A^{2} - P B^{2} = 2 k^{2}

, when

k

is constant, then find the equation to the locus of the point

P

A

and

B

being the fixed points

(a, 0)

and

(- a, 0)

respectively, obtain the equations giving the locus of

P

, when

P A^{2} - P B^{2} =

a constant quantity

= 2 k^{2}

Q. If

A (3, 0), B (- 3, 0)

then the locus of the point

P

such that

{P A}^{2} + {P B}^{2} = 18

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Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).

Let P (x, y, z) be the point if PA2+PB2=2k2 ⇒x-32+y-42+z-522+x+12+y-32+z+722=2k2⇒x2-6x+9+y2-8y+16+z2-10z+25+x2+2x+1+y2-6y+9+z2+14z+49=2k2⇒2x2+2y2+2z2-4x-14y+4z+109-2k2=0 Hence, 2x2+2y2+2z2-4x-14y+4z+109-2k2=0 is the locus of point P.

Find the locus of P if PA² + PB²= 2k², where A and B are the points (3, 4, 5) and (–1, 3, –7).