JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 2)
Let A(1, 4) and B(1, $$-$$5) be two points. Let P be a point on the circle
(x $$-$$ 1)2 + (y $$-$$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :
(x $$-$$ 1)2 + (y $$-$$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :
a straight line
an ellipse
a parabola
a hyperbola
Explanation
P be a point on $${(x - 1)^2} + {(y - 1)^2} = 1$$
so $$P(1 + \cos \theta ,1 + \sin \theta )$$
A(1, 4), B(1, $$-$$5)
$${(PA)^2} + {(PB)^2}$$
$$ = {(\cos \theta )^2} + {(\sin \theta - 3)^2} + {(\cos \theta )^2} + {(\sin \theta + 6)^2}$$
$$ = 47 + 6\sin \theta $$
It is maximum if $$\sin \theta = 1$$
When $$ \sin \theta = 1,\cos \theta = 0$$
So P(1, 2), A(1, 4), B(1, $$-$$5)
P, A, B are collinear points.
so $$P(1 + \cos \theta ,1 + \sin \theta )$$
A(1, 4), B(1, $$-$$5)
$${(PA)^2} + {(PB)^2}$$
$$ = {(\cos \theta )^2} + {(\sin \theta - 3)^2} + {(\cos \theta )^2} + {(\sin \theta + 6)^2}$$
$$ = 47 + 6\sin \theta $$
It is maximum if $$\sin \theta = 1$$
When $$ \sin \theta = 1,\cos \theta = 0$$
So P(1, 2), A(1, 4), B(1, $$-$$5)
P, A, B are collinear points.
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