Question

Draw the graph by using the following equation

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

Answer 1

Hint: When plotting any graph of curve equations first convert the given equation into standard form and find the intersection points, focus and directrix. Using this data plot the graph of the given curve equation.

Complete step by step answer:

 First find out the intersection points with x and y axis

 put x = 0 in equation

 $3{y}^2 + 4y + 1 = 0$

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

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

 Now put y = 0 in equation

 $ \Rightarrow x = 1 $

 $ \Rightarrow$ intersection point with x axis is (1,0)

 Now convert the equation into standard form

  $ \Rightarrow 3{y}^2 + 4y = x - 1$

  Divide by 3

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

Now add and subtract by $(\dfrac{4}{6})^2$ in L.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 the 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 $Y = 0 \& X = 0$

 $ \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 $(- \dfrac{1}{3}, - \dfrac{2}{3}) $

Focus: The coordinates of the focus of the parabola w.r.t standard parabola is $(X = a,Y = 0)$

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

$\Rightarrow (y + \dfrac{2}{3}) = 0 \Rightarrow y = - \dfrac{2}{3} $

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

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 we have to find the intersection point then convert into standard equation then compare it with standard equation of parabola then solve it to find the solution.