Question

Find the locus of a point which moves such that the difference of its distances from the points

(3,0) and (-3,0) is 4 units.
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Answer 1

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.

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Answer 2

Answer:

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Step-by-step explanation:

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