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Grade: 12th pass
        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?
4 months ago

## Answers : (1)

Samyak Jain
329 Points
							A(0,4) , B(2t,0) , M is mid-point of AB. So, by mid-point formula.M $\dpi{100} \equiv$ ((0+2t)/2, (4+0)/2) $\dpi{100} \equiv$ (t,2).Slope of AB is (0 – 4)/(2t – 0) = – 2/t = m1 (let)$\dpi{100} \therefore$ 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 isy – 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) / 2So, 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)  $\dpi{100} \Rightarrow$ h2 = t2/4   or   t2 = 4h2Substitute the value of t2 in (3).k = (8 – 4h2) / 4  $\dpi{100} \Rightarrow$  k = 2 – h2Replace h by x and k by y to get locus of point P which isy = 2 – x2  i.e.  y + x2 = 2.Ans : (A).

4 months ago
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