Find the locus of a point whose difference of distance from points (3,00 and (-3,0) is 4.
Answers
Step-by-step explanation:
▶let the p (x, y) be a point on the locus ,
▶le us consider the points are
A (3,0)
B(-3, 0)
so the condition given is ,
AP - BP = 4
AP = 4 + BP
use the distance formula
√[(y - 0)² + (x - 3)²]
= 4 + √[(y - 0 )² + (x -(-3))²]
squaring on both the sides
(y - 0)² + (x - 3)²
= ( (4)² + 2 x 4 x √[(y - 0 )² + (x -(-3))²] + ((y - 0)² + (x + 3)²)
((y - 0)² + (x - 3)²) - ((y - 0)² + (x + 3)²) = ( (4)² + 2 x 4 x √[(y - 0 )² + (x -(-3))²]
( x - 3 )² - ( x + 3)²= ( (4)² + 2 x 4 x √[(y - 0 )² + (x -(-3))²]
( x - 3 )² - ( x + 3)²=( 16 + 8 √[(y - 0 )² + (x -(-3))²] )
(x² + 6x + 9 ) - (x² - 6x + 9 )= 16 + 8 √[(y - 0 )² + (x -(-3))²]
(x² + 6x + 9 - x² + 6x - 9 ) = 16 + 8 √[(y - 0 )² + (x -(-3))²]
6x + 6x = 16 + 8 √[(y - 0 )² + (x -(-3))²]
12x - 16 = 8 √[(y - 0 )² + (x -(-3))²]
4 ( 3x - 4 ) = 8 √[(y - 0 )² + (x -(-3))²]
( 3x - 4) = 2 √[(y - 0 )² + (x -(-3))²]
squaring both the sides
9x² - 2 (3x) (4) + 16 = 4 ((y - 0 )² + (x -(-3))²)
9x² - 8 (3x) + 16 = 4 ((y - 0 )² + (x -(-3))²)
9x² - 24 x + 16= 4 (( y² - 0 + 0 ) + (x²- 2(x)(-3)+ 9))
9x² - 24x + 16 = 4 (y² + x² + 6x + 9 )
9x² - 24x + 16 = 4y² + 4x² + 24x + 36
9x² - 4x² - 4y² + 16 - 36
5x² - 4y² - 20 = 0
5x² - 4y² = 20
divide by 20 on the both sides
(x² / 4) - (x² / 5 ) = 1
(x² / 4) - (x² / 5 ) = 1 is the locus of a point whose difference of distance from points (3,0)and (-3,0) is 4.