WBJEE · Maths · Straight Lines
Let \(A\) be the point \((0,4)\) and \(B\) be a moving point on \(x\)-axis. Let \(M\) be the midpoint of \(A B\) and let the perpendicular bisector of \(A B\) meets the \(y\)-axis at \(R\). The locus of the midpoint \(P\) of \(M R\) is
- A \(y+x^{2}=2\)
- B \(x^{2}+(y-2)^{2}=\frac{1}{4}\)
- C \((y-2)^{2}-x^{2}=\frac{1}{4}\)
- D \(x^{2}+y^{2}=16\)
Answer & Solution
Correct Answer
(A) \(y+x^{2}=2\)
Step-by-step Solution
Detailed explanation
Let B \(=(2 \alpha, 0)\) \(\therefore \mathrm{M}=(\alpha, 2)\) \(\mathrm{m}_{\mathrm{AB}}=\frac{4}{-2 \alpha}=-\frac{2}{\alpha}\) \(\therefore \mathrm{m}_{\mathrm{MR}}=\frac{\alpha}{2}, \therefore \mathrm{MR} \quad \mathrm{y}-2=\frac{\alpha}{2}(\mathrm{x}-\alpha)\) Put,…
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