Find the equation of a locus of a point which moves so that the sum of its distance from (2,0)and (-2,0) is 8.
it is about equation of a locus..
please help!!
Your description defines an ellipse.
does the question expect you to find that equation by using that definition?
if so, then let such a point on that locus be P(x,y)
√[(x-2)^2 + y^2] + √[(x+2)^2 + y^2] = 8
√[(x-2)^2 + y^2] = 8 - √[(x+2)^2 + y^2]
square both sides and expand
x^2 - 4x + 4 + y^2 = 64 - 16√[(x+2)^2 + y^2] + x^2 + 4x + 4 + y^2
16√[(x+2)^2 + y^2] = 64 + 8x
2√[(x+2)^2 + y^2] = 8 + x
squaring again and simplifying I get
3x^2 + 4y^2 = 48
divide each term by 48 to get it into standard form
x^2/16 + y^2/12 = 1
or, the easy way
from the description 2a = 8 , a = 4
(2,0) and (-2,0) must be the focal points so the midpoint or (0,0) must be the centre and c = 2
since the focal points lie on the x-axis
b^2 + c^2 = a^2
b^2 + 4 = 16
b^2 = 12
standard form with centre (0,0) is
x^2/a^2 + y^2/b^2 = 1
so
x^2/16 +y^2/12 = 1
To find the equation of the locus of a point, we need to determine the relationship that satisfies the given condition. In this case, we are looking for the locus of a point such that the sum of its distances from (2,0) and (-2,0) is 8.
Let's assume the coordinates of the point are (x, y). We can start solving the problem by applying the distance formula:
Distance between (x, y) and (2,0): √[(x - 2)² + (y - 0)²]
Distance between (x, y) and (-2,0): √[(x + 2)² + (y - 0)²]
According to the condition given, the sum of these two distances should be 8. So, we can form the equation:
√[(x - 2)² + (y - 0)²] + √[(x + 2)² + (y - 0)²] = 8
This equation represents the locus of points that satisfy the given condition.