JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 22)
If a point P has co-ordinates (0, $$-$$2) and Q is any point on the circle, x2 + y2 $$-$$ 5x $$-$$ y + 5 = 0, then the maximum value of (PQ)2 is :
$${{25 + \sqrt 6 } \over 2}$$
14 + $$5\sqrt 3 $$
$${{47 + 10\sqrt 6 } \over 2}$$
8 + 5$$\sqrt 3 $$
Explanation
Given that x2 + y2 $$-$$ 5x $$-$$ y + 5 = 0
$$ \Rightarrow $$ (x $$-$$ 5/2)2 $$-$$ $${{25} \over 4}$$ + (y $$-$$ 1/2)2 $$-$$ 1/4 = 0
$$ \Rightarrow $$ (x $$-$$ 5/2)2 + (y $$-$$ 1/2)2 = 3/2
on circle [ Q $$ \equiv $$ (5/2 + $$\sqrt {3/2} $$ cos Q, $${1 \over 2}$$ + $$\sqrt {3/2} $$ sin Q)]
$$ \Rightarrow $$ PQ2 = $${\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2}$$ + $${\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}$$
$$ \Rightarrow $$ PQ2 = $${{25} \over 2} + {3 \over 2} + 5\sqrt {3/2} $$ (cos Q + sinQ)
= 14 + 5$$\sqrt {3/2} $$ (cosQ + sinQ)
$$ \therefore $$ Maximum value of PQ2
= 14 + 5$$\sqrt {3/2} $$ $$ \times $$ $$\sqrt 2 $$ = 14 + 5$$\sqrt 3 $$
$$ \Rightarrow $$ (x $$-$$ 5/2)2 $$-$$ $${{25} \over 4}$$ + (y $$-$$ 1/2)2 $$-$$ 1/4 = 0
$$ \Rightarrow $$ (x $$-$$ 5/2)2 + (y $$-$$ 1/2)2 = 3/2
on circle [ Q $$ \equiv $$ (5/2 + $$\sqrt {3/2} $$ cos Q, $${1 \over 2}$$ + $$\sqrt {3/2} $$ sin Q)]
$$ \Rightarrow $$ PQ2 = $${\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2}$$ + $${\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}$$
$$ \Rightarrow $$ PQ2 = $${{25} \over 2} + {3 \over 2} + 5\sqrt {3/2} $$ (cos Q + sinQ)
= 14 + 5$$\sqrt {3/2} $$ (cosQ + sinQ)
$$ \therefore $$ Maximum value of PQ2
= 14 + 5$$\sqrt {3/2} $$ $$ \times $$ $$\sqrt 2 $$ = 14 + 5$$\sqrt 3 $$
Comments (0)
