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  • If parametric representation of a parabola is  <img alt="\mathrm {x=2+t^2 \: \: \&\: \: y=2 t+1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7Bx%3D2&plus;t%5E2%2

If parametric representation of a parabola is  \mathrm {x=2+t^2 \: \: \&\: \: y=2 t+1} , then
 

Option: 1

axis of parabola is \mathrm y=1
 


Option: 2

equation of directrix is \mathrm x=1
 


Option: 3

focus of parabola is  \mathrm S(3,1)
 


Option: 4

All of the above


Answers (1)

best_answer

Given 

\begin{aligned} &\mathrm {x=2+t^2 \& y=2 t+1} \\ & \mathrm {\Rightarrow t^2=x-2 \& t^2=\left(\frac{y-1}{2}\right)^2} \\ &\mathrm { \therefore \quad \frac{(y-1)^2}{4}=x-2} \end{aligned}

\mathrm {\Rightarrow(y-1)^2=4(x-2) \: \: or\: \: (y-1)^2=4(1)(x-2)}\\

\mathrm {\Rightarrow Y^2=4(1) X \: \: where\: \: X=x-2, Y=y-1}


\therefore  Vertex is \mathrm {V(X=0, Y=0)=V(2,1)} 


Focus is  \mathrm {S(X=1, Y=0)} 


\mathrm { =S(x-2=1, y-1=0)=S(3,1) }


Again equation of directrix is  \mathrm { x-2=-1 \Rightarrow x=1 } 

and axis of parabola is

\mathrm { y-1=0 \Rightarrow y=1 }

Posted by

Divya Prakash Singh

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