Question

find the equation of locus of a point,the difference of whose distances from (-5,0) and (5,0) is 8.

Answer 1

Let coordinates of P (h,k), Acordingly:-

PA+PB=6

PA= 6-PB

or. PA^2=36+PB^2-12PB

or. 12.PB=36+PB^2-PA^2

or. 12.PB=36+h^2+(k+2)^2-h^2-(k-2)^2

or. 12.PB=36+8k

or. 3.PB=9+2k

or. 9.PB^2=81+36k+4k^2

or. 9[(h-0)^2+(k+2)^2]=81+36k+4k^2

or. 9h^2+9k^2+36k+36 =81+36k+4k^2

or. 9h^2+5k^2=45

Locus of (h,k) is:-

9x^2+5y^2= 45. Answer

Answer 2

Let the point be (x,y),

By using distance formula for coordinates,

[(x+5)

2

+(y−0)

2

]

2

1

−[(x−5)

2

+(y−0)

2

]

2

1

=8

⇒[x

2

+10x+25+y

2

]

2

1

=[x

2

−10x+25+y

2

]

2

1

+8

squaring on both sides,

⇒x

2

+10x+25+y

2

=x

2

−10x+25+y

2

+64+16[x

2

−10x+25+y

2

]

2

1

⇒5x−16=4[x

2

−10x+25+y

2

]

2

1

Again squaring both sides,

⇒25x

2

+256−160x=16x

2

−160x+400+16y

2

Therefore, locus is,