Find the equation of the locus of a point, the difference of whose distances from (-5,0) and
5,0) is 8.
Answers
Step-by-step explanation:
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,
9x
2
−16y
2
=144.......(hyperbola)
Answer:
144=(3x-4y) (3x+4y)
Step-by-step explanation:
let the point by (x, y)
so, √(x-(-5))^2+(y-0)^2-√(x-5)^2+(y-0)^2=8
so √(x+5)^2+(y)^2 - 8 =√(x-5)^2+(y)^2
so √x^2+25+10x+y^2-8=√x^2+25-10x+y^2
so squaring both sides we will get
=> x^2+25+10x+y^2+64+16√(x^2+25+10x)+y^2
=x^2+25-10x+y^2
cancelling out terms that can be cancelled
10x+64+16√(x+5)^2+y^2= -10x
20x+16√(x+5)^2+y^2= -64
taking 4 common and cancelling on both sides
5x+4√(x+5)^2+y^2= -16
√(x+5)^2+y^2 = -(16+5x)/4
(x+5)^2+y^2= (256+25x^2+160x/)16
16(x^2+25+10x+y^2)=256+25x^2+160x
16x^2+400+160x+16y^2=256
144=(3x+4y)(3x-4y)
this is the locus as far as i solve.
well, i may have made some mistakes because i have not paper solved ut, but i have rechecked and see no mistakes.