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Draw the graph by using the following equation:

$  x = 3{y^2} + 4y + 1 $


Answer
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First find out the intersection points with x and y axis

put x = 0 in equation

  ${\text{3}}{{\text{y}}^2} + 4y + 1 = 0 $

$   \Rightarrow y = - \dfrac{1}{3},\; - 1 $

$   \Rightarrow $ Intersection point with y axis is $\left( {0, - \dfrac{1}{3}} \right)\,\left( {0, - 1} \right) $

Now put y = 0 in equation

  $ \Rightarrow x = 1 $

  $ \Rightarrow $ Intersection point with x axis is $\left( {1,0} \right)$

Now convert the equation into standard form

 $ {\text{3}}{{\text{y}}^2} + 4y = x - 1 $

  ${\text{Divide by 3}} $

  ${{\text{y}}^2} + \dfrac{4}{3}y = \dfrac{1}{3}x - \dfrac{1}{3}$

  Now add and subtract by ${\left( {\dfrac{4}{6}} \right)^2}{\text{ in L}}{\text{.h}}.s$  to make a complete square in y

 $  \Rightarrow {{\text{y}}^2} + \dfrac{4}{3}y + {\left( {\dfrac{4}{6}} \right)^2} - {\left( {\dfrac{4}{6}} \right)^2} = \dfrac{1}{3}x - \dfrac{1}{3} $

$   \Rightarrow {\left( {y + \dfrac{2}{3}} \right)^2} = \dfrac{1}{3}x - \dfrac{1}{3} + \dfrac{4}{9} $

 $  \Rightarrow {\left( {y + \dfrac{2}{3}} \right)^2} = \dfrac{1}{3}x + \dfrac{1}{9} $

  $ \Rightarrow {\left( {y + \dfrac{2}{3}} \right)^2} = \dfrac{1}{3}\left( {x + \dfrac{1}{3}} \right) $

   So, you know this is the equation of horizontal parabola opening right side

  Now compare it with standard equation ${Y^2} = 4aX $

   $\Rightarrow Y = \left( {y + \dfrac{2}{3}} \right)$  &  $X = \left( {x + \dfrac{1}{3}} \right)$ & $ 4a = \dfrac{1}{3} $     $\Rightarrow a = \dfrac{1}{{12}} $

  Vertex of the standard parabola is Y = 0 and X=0

$ \Rightarrow \left( {y + \dfrac{2}{3}} \right)= 0$  &  $ \left( {x + \dfrac{1}{3}} \right) = 0 $

   $\Rightarrow y = - \dfrac{2}{3}$ & $ x = - \dfrac{1}{3} $


  So, vertex of the given parabola is $\left( { - \dfrac{1}{3}\;, - \dfrac{2}{3}} \right)\; $


Focus: the coordinates of the focus of the parabola w ${\text{.r}}{\text{.t standard parabola is }}\left( {X = a,\;Y = 0} \right) \\$

$  X = \dfrac{1}{{12}}\;\; \Rightarrow {\text{ }}\left( {x + \dfrac{1}{3}} \right) = \dfrac{1}{{12}}\;\; \Rightarrow x = - \dfrac{1}{4} $

 $ \left( {y + \dfrac{2}{3}} \right)\; = 0\,\; \Rightarrow y = - \dfrac{2}{3} \\$

So, coordinates of focus of your given equation is $\left( { - \dfrac{1}{4}, - \dfrac{2}{3}} \right) $

  $Directrix:\;$ The equation of directrix w.r.t standard parabola is  $X = - a $

  $ \Rightarrow \;\;X = - \dfrac{1}{{12}} $

  So, the equation of directrix of your given equation is:

  $ \Rightarrow x + \dfrac{1}{3} = - \dfrac{1}{{12}}\; \Rightarrow x = - \dfrac{5}{{12}} $

NOTE: - In this type of question first find out the intersection point then convert into standard equation then compare it from standard equation of parabola then solve it you will get your answer.