A) a pair of straight lines
B) an ellipse
C) a parabola
D) a hyperbola
Correct Answer: C
Solution :
We have, \[x={{t}^{2}}+t+1\] ... (i) and \[y={{t}^{2}}-t+1\] ... (ii) Now \[x+y=2(1+{{t}^{2}})\] ... (iii) and \[x-y=2t\] ... (iv) Now, from Eqs. (iii) and (iv), we get \[x+y=2\left[ 1+{{\left( \frac{x-y}{2} \right)}^{2}} \right]\] \[\Rightarrow \] \[x+y=2\left[ \frac{4+{{x}^{2}}+y\,\,\,\,\,\,\,2xy}{4} \right]\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2xy-2x-2y+4=0\] ... (v) \[\therefore \]On comparing with \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\] We get,\[a=1,\,\,b=1,\,\,c=4,\,\,h=-1,\,\,g=-1,\,\,f=-1\] \[\because \] \[\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}\] Now, \[\Delta 1\cdot 1\cdot 4+2(-1)(-1)(-1)-1\times {{(-1)}^{2}}\] \[=4-2-1-1-4\] \[=-4\]therefore,\[\Delta \ne 0\] and \[ab-{{h}^{2}}=1\cdot 1-{{(1)}^{2}}=1-1=0\] So, it is equation of a parabola.
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