Question

Find the equation of locus of a point P, the square of whose distance from the origin is 4 times its
y-coordinate.??​

Answer 1

=》x^2+y^2-4y=0. Therefore Equation of locus of P is x^2+y^2-4y=0.

hope it helps you

mark as brainest and follow

Answer 2

$\huge\mathcal{\fcolorbox{aqua}{azure}{\red{♡Answer★}}}$

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:

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

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

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

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

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

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

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

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