Question

P is a point which moves in the xy plane such that the point p is nearer to the centre of a square than any of the sides. the four vertices of the square are

Answer 1

question is incomplete. A complete question is -----> P is a point which moves in the xy plane such that the point p is nearer to the centre of a square than any of the sides. the four vertices of the square are (±a, ±a). The region in which P will move is bounded by part of parabola/parabolas having equations ?


solution :- see figure, Let ABCD is a square, where origin O is the centre of square and P(x,y) lies inside in such a way that OP < PM
OP < PR , OP < PN and OP < PQ.

here, A(a, a) , B(-a, a) , C(-a, -a) and D (a , -a) and P(x, y) then, PM = |a - x| , PR = |a + x|
PN = |a - y| and PQ = |a + y|

for OP < PM
$\sqrt{x^2+y^2}<|a-x|$
squaring both sides,
x² + y² = (a - x)²
x² + y² = a² + x² - 2ax
y² = x² - 2ax


similarly, we get,
x² + y² = (a + x)² => y² = x² + 2ax
x² + y² = (a - y)² => x² = y² - 2ay
x² + y² = (a + y)² => x² = y² + 2ay

Answer 2

Answer:

The above answer is great.