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Let A be the fixed point (0, 4) and B be a moving point (2t, 0). Let M be the mid-point of AB and let the perpendicular bisector of AB meet the y-axis at R. The locus of the mid-point P of MR is (A) y + x 2 = 2 (B) x 2 + (y − 2)2 = 1/4 (C) (y − 2)2 − x 2 = ¼ (D) none of the above?

Let A be the fixed point (0, 4) and B be a moving point (2t, 0). Let M be the mid-point of AB and let the perpendicular bisector of AB meet the y-axis at R. The locus of the mid-point P of MR is (A) y + x 2 = 2 (B) x 2 + (y − 2)2 = 1/4 (C) (y − 2)2 − x 2 = ¼ (D) none of the above?
 

Grade:12th pass

1 Answers

Samyak Jain
333 Points
5 years ago
A(0,4) , B(2t,0) , M is mid-point of AB. So, by mid-point formula.
M \equiv ((0+2t)/2, (4+0)/2) \equiv (t,2).
Slope of AB is (0 – 4)/(2t – 0) = – 2/t = m1 (let)
\therefore Slope (m2) of perpendicular bisector of AB is given by m1m2 = – 1.
We get m2 = t/2.
Equation of perpendicular bisector of AB using slope(t/2) - point(M) form is
y – 2 = (t/2)(x – t)                   ...(1)
To get the coordinates of R which lies on this line and y-axis, put x = 0 in eq(1).
y – 2 = (t/2)(0 – t) = – t2 / 2  or  y = 2 – t2 / 2  =  (4 – t2) / 2
So, R(0 , (4 – t2) / 2).
Now, mid-point of MR is P((t + 0)/2 , (2 + (4 – t2) / 2)/2)  i.e. P(t/2 , (8 – t2)/4)  [Mid-point formula]
We need to eliminate parameter t to find the locus of P.
Let coordinates of P be (h , k). Then h = t/2   ...(2) and k = (8 – t2)/4    ...(3)
Square both sides of (2)  \Rightarrow h2 = t2/4   or   t2 = 4h2
Substitute the value of t2 in (3).
k = (8 – 4h2) / 4  \Rightarrow  k = 2 – h2
Replace h by x and k by y to get locus of point P which is
y = 2 – x2  i.e.  y + x2 = 2.
Ans : (A).

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