A) \[{{y}^{2}}=2x+8\]
B) \[{{y}^{2}}=x+8\]
C) \[{{y}^{2}}=2x-8\]
D) \[{{y}^{2}}=x-8\]
Correct Answer: C
Solution :
[c] Given parabola is \[{{y}^{2}}=4x\] ? (1) Let \[P\equiv \left( {{t}^{2}}_{1},2{{t}_{1}} \right)\] and \[Q\equiv \left( {{t}^{2}}_{2},2{{t}_{2}} \right)\] Slope of \[OP=\frac{2{{t}_{1}}}{{{t}^{2}}_{1}}=\frac{2}{{{t}_{1}}}\] and slope of \[OQ=\frac{2}{{{t}_{2}}}\] Since \[OP\bot OQ,\] \[\therefore \frac{4}{{{t}_{1}}{{t}_{2}}}=-1\] or \[{{t}_{1}}{{t}_{2}}=-4\] ? (2) Let \[R(h,k)\] be the middle point of PQ, then \[h=\frac{t_{1}^{2}\,\,+\,\,t_{2}^{2}}{2}\] ? (3) and \[k={{t}_{1}}+{{t}_{2}}\] ? (4) From (4), \[{{k}^{2}}={{t}^{2}}_{1}+{{t}^{2}}_{2}+2{{t}_{1}}{{t}_{2}}=2h-8\] [From (2) and (3)] Hence locus of \[R(h,k)\] is \[{{y}^{2}}-2x-8.\]
You need to login to perform this action.
You will be redirected in
3 sec
