A) \[{{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\]
B) \[{{t}_{2}}=-{{t}_{1}}+\frac{2}{{{t}_{1}}}\]
C) \[{{t}_{2}}={{t}_{1}}-\frac{2}{{{t}_{1}}}\]
D) \[{{t}_{2}}={{t}_{1}}+\frac{2}{{{t}_{1}}}\]
Correct Answer: A
Solution :
[a] Given point on parabola is \[P(b{{t}_{1}}^{2},2b{{t}_{1}})\] So consider parabola \[{{y}^{2}}=4bx\] Differentiating W.r.t.x, we get \[2y\frac{dy}{dx}=4b\] \[\therefore \frac{dy}{dx}=\frac{2b}{y}\] \[\therefore \]slope of normal at point \[p=-\frac{2b{{t}_{1}}}{2b}=-{{t}_{1}}\]. Also normal meets the curve again at point \[Q(b{{t}^{2}}_{2},2b{{t}_{2}})\] So slope of line \[PQ=\frac{2b{{t}_{2}}-2btl1}{b{{t}^{2}}_{2}-b{{t}^{2}}_{1}}=\frac{2}{{{t}_{1}}+{{t}_{2}}}\] \[\therefore -{{t}_{1}}=\frac{2}{{{t}_{1}}+{{t}_{2}}}\] \[\therefore {{t}_{2}}=-{{t}_{1}}-\frac{2}{{{t}_{1}}}\]
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