JEE Mains · Maths · STD 11 - 9. straight line
Let \(\mathrm{A}\) be a fixed point \((0,6)\) and \(\mathrm{B}\) be a moving point \((2 \mathrm{t}, 0)\). Let \(\mathrm{M}\) be the mid-point of \(\mathrm{AB}\) and the perpendicular bisector of \(\mathrm{AB}\) meets the \(\mathrm{y}\)-axis at \(\mathrm{C}\). The locus of the mid-point \(\mathrm{P}\) of \(\mathrm{MC}\) is :
- A \(3 x^{2}-2 y-6=0\)
- B \(3 x^{2}+2 y-6=0\)
- C \(2 \mathrm{x}^{2}+3 \mathrm{y}-9=0\)
- D \(2 \mathrm{x}^{2}-3 \mathrm{y}+9=0\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{x}^{2}+3 \mathrm{y}-9=0\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}(0,6)\) and \(\mathrm{B}(2 \mathrm{t}, 0)\) Perpendicular bisector of \(\mathrm{AB}\) is \((y-3)=\frac{t}{3}(x-t)\) So, \(C=\left(0,3-\frac{t^{2}}{3}\right)\) Let \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\)…
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