Let A be a fixed point (0,6) and B be a moving point (2 t , 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 :

Question

Let A be a fixed point (0,6) and B be a moving point (2 t , 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 : (A) 3 x2-2 y-6=0 (B) 3 x2+2 y-6=0 (C) 2 x2+3 y-9=0 (D) 2 x2-3 y+9=0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/58aa304b07f2bd5dd2d5df163202386d-.png" /> Perpendicular bisector of AB is (y-3)=(t/3)(x-t) So, C=(0,3-(t2/3)) Let P be ( h , k ) h =( t /2) ; k =(3-( t 2/6)) ⇒ k =3-(4 h 2/6) ⇒ 2 x 2+3 y -9=0

Q. Let $A$ be a fixed point $(0, 6)$ and $B$ be a moving point $(2 t, 0)$ . Let $M$ be the mid-point of $A B$ and the perpendicular bisector of $A B$ meets the $y$ -axis at $C$ . The locus of the mid-point $P$ of $MC$ is :

$3 x^{2} - 2 y - 6 = 0$

$3 x^{2} + 2 y - 6 = 0$

$2 x^{2} + 3 y - 9 = 0$

$2 x^{2} - 3 y + 9 = 0$

Perpendicular bisector of $A B$ is
$(y - 3) = \frac{t}{3} (x - t)$
So, $C = (0, 3 - \frac{t ^{2}}{3})$
Let $P$ be $(h, k)$
$h = \frac{t}{2}; k = (3 - \frac{t ^{2}}{6})$
$\Rightarrow k = 3 - \frac{4 h ^{2}}{6}$
$\Rightarrow 2 x^{2} + 3 y - 9 = 0$

Q. 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