JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 1)
Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
3x2 $$-$$ 2y $$-$$ 6 = 0
3x2 + 2y $$-$$ 6 = 0
2x2 + 3y $$-$$ 9 = 0
2x2 $$-$$ 3y + 9 = 0
Explanation
A(0, 6) and B(2t, 0)
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Perpendicular bisector of AB is
$$(y - 3) = {t \over 3}(x - t)$$
So, $$C = \left( {0,3 - {{{t^2}} \over 3}} \right)$$
Let P be (h, k)
$$h = {t \over 2};k = \left( {3 - {{{t^2}} \over 6}} \right)$$
$$ \Rightarrow k = 3 - {{4{h^2}} \over 6} \Rightarrow 2{x^2} + 3y - 9 = 0$$
Perpendicular bisector of AB is
$$(y - 3) = {t \over 3}(x - t)$$
So, $$C = \left( {0,3 - {{{t^2}} \over 3}} \right)$$
Let P be (h, k)
$$h = {t \over 2};k = \left( {3 - {{{t^2}} \over 6}} \right)$$
$$ \Rightarrow k = 3 - {{4{h^2}} \over 6} \Rightarrow 2{x^2} + 3y - 9 = 0$$
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