Man running on a race track notes that the sum of the distances of two flag posts from him is always 10 metres and the distance between the flag posts is 8 metres. The area of the path he encloses (in square metres) is
Answers
Correct Question :
A man running on a race track note that the sum of the distances from the two flag posts form him is always 10 m and the distance between the flag posts is 8 m. Find the area of the path he encloses.
Answer :
Given
Refer to attachment for graph
A man is running a race course.
Sum of the distances from the two flag posts from him is always 10, which means it is constant
Distance between the flag posts is 8 m
By the given question the equation of the posts traced by the man is ellipse.
Here the coordinates of the flag posts are foci
Coordinates of the center = ( 0, 0 )
Distance between the flag posts = 8 m
Nothing but distance between foci = 8 m
SS' = 8 m
SC + CS' = 8 m
2a = 8
a = 8/2 = 4
Coordinates of the foci ( ± a, 0 ) = ( ± 4, 0 )
S = ( 4, 0 )
S' = ( - 4, 0 )
Sum of the distance from the flag posts from him = 10 m
PS + PS' = 10
√[ ( x - 4 )² + y² ] + √[ ( x + 4 )² + y² ] = 10
√[ x² + y² - 8x + 16 ] + √[ x² + y² + 8x + 16 ] = 10
√[ x² + y² - 8x + 16 ] = 10 - √[ x² + y² + 8x + 16 ]
Squaring on both sides
x² + y² - 8x + 16 = 10² + x² + y² + 8x + 16 - 20√[ x² + y² + 8x + 16 )
- 16x - 100 = - 20√[ x² + y² + 8x + 16 ]
4x + 25 = 5√[ x² + y² + 8x + 16 ]
( 4x + 25 )² = 25( x² + y² + 8x + 16 )
16x² + 625 + 200x = 25x² + 200x + 5²y² + 400
625 - 400 = 3²x² + 5²y²
225 = 3²x² + 5²y²
15² = 3²x² + 5²y²
1 = 3²x²/15² + 5²y²/15²
1 = x²/5² +y²/3²
x²/5² + y²/3² = 1
So, b = 3
Area of the path he encloses = Area of the ellipse
= πab
= π( 5 ) ( 3 )
= 15π m²
Hence the area of path the man enclosed is 15π m².
