A) \[({{t}^{2}}+3,-{{t}^{3}}-1)\]
B) \[({{t}^{2}}+3,{{t}^{3}}-1)\]
C) \[(16{{t}^{2}}+3,-64{{t}^{3}}-1)\]
D) \[(4{{t}^{2}}+3,-4{{t}^{3}}-1)\]
Correct Answer: A
Solution :
\[P(x=4{{t}^{2}}+3,y=8{{t}^{2}}-1)\] let\[Q(4t_{1}^{2}+3,8y_{1}^{3}-1)\] at\[P,\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{24{{t}^{2}}}{8t}=3t\] \[\therefore \]tangent at P is\[y-8{{t}^{3}}+1=3t(x-4{{t}^{2}}-3)\] Q will satisfy it \[\therefore \]\[8t_{1}^{3}-8{{t}^{3}}=3t(4t_{1}^{2}-4{{t}^{2}})\] \[8({{t}_{1}}-t)(t_{1}^{2}+{{t}_{1}}t+{{t}^{2}})=3t.4({{t}_{1}}-t)({{t}_{1}}+t)\] \[2(t_{1}^{2}+{{t}_{1}}t+{{t}^{2}})=3t({{t}_{1}}+t)\] \[2t_{1}^{2}+2{{t}_{1}}t+2{{t}^{2}}=3t{{t}_{1}}+3{{t}^{2}}\] \[2t_{1}^{2}-{{t}_{1}}t-{{t}^{2}}=0\] \[({{t}_{1}}-t)(2{{t}_{1}}+t)=0\] \[{{t}_{1}}=-\frac{t}{2}\] \[\therefore \]\[Q({{t}^{2}}+3,-{{t}^{3}}-1)\]Ans. (a)
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