question_answer

A line PQ meets the parabola \[{{y}^{2}}-4ax\] in R such that PQ is bisected at R. if the coordinates of P are \[({{x}_{1}},{{y}_{1}})\] then the locus of Q is the parabola

A) \[{{(y+{{y}_{1}})}^{2}}=8a(x+{{x}_{1}})\]

B) \[{{(y-{{y}_{1}})}^{2}}=8a(x+{{x}_{1}})\]

C) \[{{(y+{{y}_{1}})}^{2}}=8a(x-{{x}_{1}})\]

D) None of these

Correct Answer: A

Solution :

[a] Let the coordinates of Q be\[(h,k)\]. Since the point R lies on the parabola. Let it coordinates be \[(a{{t}^{2}},2at).\] Since R is midpoint of PQ, \[\therefore \,\,\,\,a{{t}^{2}}=\frac{{{x}_{1}}+h}{2}\] and \[2at=\frac{{{y}_{1}}+k}{2}\] \[\Rightarrow {{t}^{2}}=\frac{{{x}_{1}}+h}{2a}\] And \[t=\frac{{{y}_{1}}+k}{4a}\] Equating the two values of t, we get \[{{\left( \frac{{{y}_{1}}+k}{4a} \right)}^{2}}=\frac{{{x}_{1}}+h}{2a}\Rightarrow {{({{y}_{1}}+k)}^{2}}=8a({{x}_{1}}+h)\] Hence, locus of Q (h, k) is \[{{(y+{{y}_{1}})}^{2}}=8a(x+{{x}_{1}})\]