Question

Find equation to the locus of the point ,the square of whose distance from origin is 4 times its y-coordinate

Answer 1

given
$\sqrt{ x^{2} + y^{2} } i$
is the distance from origin to the point (x,y)
given that
$( \sqrt{ x^{2} + y^{2} }) ^{2} =4y$
$x^{2} + y^{2} =4y$
is the required equation of locus

Answer 2

Answer:

The equation of locus of the point, the distance of whose distance from origin is 4 times its y-coordinate is x² + y² - 4y = 0 .

Step-by-step explanation:

• A locus is a set of points which satisfy certain geometric conditions.
• Distance between two points ( x1, y1 ) and ( x2 , y2 ) is given by:

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

• Distance of a point ( x , y ) from origin is given by:

$\sqrt{ {x}^{2} + {y}^{2} }$

Given that :

• Square of distance of point from origin is equal to 4 times its y-coordinate.

Solution:

• Let, the given point be A( x , y ). Then, distance from this point from origin ( 0 , 0 ) is

$d = \sqrt{ {x}^{2} + {y}^{2} }$

• The square of distance, d is equal to 4 times the 4 times its y-coordinate which gives:

${d}^{2} = 4y \\ {x}^{2} + {y}^{2} = 4y \\ {x}^{2} + {y}^{2} - 4y = 0$

• Hence, the equation of locus is x² + y² - 4y = 0 .