A) 1
B) 2
C) \[\frac{1}{2}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
The normal of P(t) to \[{{y}^{2}}=8x\] is \[xt+y=2{{t}^{3}}+4t\]. The point (6, 2) lies on it. \[\therefore \] \[{{t}^{3}}-t-t=0\] If \[{{t}_{1}},\,\,{{t}_{2}}\] and \[{{t}_{3}}\] are roots. \[\left. \begin{matrix} {{t}_{1}}+{{t}_{2}}+{{t}_{3}}=0,\,\,{{t}_{1}}{{t}_{2}}+{{t}_{2}}{{t}_{3}}+{{t}_{3}}{{t}_{1}}=-1 \\ \text{and}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{t}}_{1}}{{t}_{2}}{{t}_{3}}=1 \\ \end{matrix} \right\}\]?(i) Let The circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] meets the parabola at the point t. \[\therefore \] \[4{{t}^{4}}+16{{t}^{2}}+4g{{t}^{2}}+8ft+c=0\] \[\sum{{{t}_{1}}{{t}_{2}}{{t}_{3}}}=-2t\] and \[\sum{{{t}_{1}}{{t}_{2}}}=\frac{(16+36)}{4}=4+9\] \[\therefore \] \[{{t}_{4}}=0=c,\] \[g=-5,\,\,f=-\frac{1}{2}\] [using Eq. (i)] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-10x-y=0\] Its intercept on the y-axis is 1.
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