Question

How do you write the equation in vertex form $x=3{{y}^{2}}+5y-9$ ?

Answer 1

The above given problem is a very simple problem of parabola, and for easy solving such problems we need to remember all the formulae and different relations required for this sum. There are four general types of parabola and for each of them, the corresponding vertex and foci are as follows,

Equation	Vertex	Foci
${{y}^{2}}=4ax$	$\left( 0,0 \right)$	$\left( a,0 \right)$
${{y}^{2}}=-4ax$	$\left( 0,0 \right)$	$\left( -a,0 \right)$
${{x}^{2}}=4by$	$\left( 0,0 \right)$	$\left( 0,b \right)$
${{x}^{2}}=-4by$	$\left( 0,0 \right)$	$\left( 0,-b \right)$

Complete step by step answer:
However, the most generalised form of parabola that we can represent, given that we have the vertex of the parabola and the latus rectum is ${{\left( y-k \right)}^{2}}=4a\left( x-h \right)$ . Here in this equation, we have, $\left( h,k \right)$ as the vertex of the parabola and ‘4a’ as the length of the latus rectum. Now to write any equation of parabola in the vertex form we need to rearrange the terms of the equation in such a manner that it comes out similar to that of the generalised form.
Now we start off with the solution to the given problem by rearranging the terms of the equation in such a manner that it leads to the square of one term. We do,

& x=3{{y}^{2}}+5y-9 \\\ & \Rightarrow {{y}^{2}}+\dfrac{5}{3}y-3=\dfrac{x}{3} \\\ & \Rightarrow {{y}^{2}}+2\times \dfrac{5}{6}y+\dfrac{25}{36}-3-\dfrac{25}{36}=\dfrac{x}{3} \\\ & \Rightarrow {{y}^{2}}+2\times \dfrac{5}{6}y+\dfrac{25}{36}=\dfrac{x}{3}+3+\dfrac{25}{36} \\\ & \Rightarrow {{\left( y+\dfrac{5}{6} \right)}^{2}}=\dfrac{x}{3}+\dfrac{133}{36} \\\ & \Rightarrow {{\left( y+\dfrac{5}{6} \right)}^{2}}=\dfrac{1}{3}\left( x+\dfrac{133}{12} \right) \\\ \end{aligned}$$ From this we can very easily say that $$\left( -\dfrac{133}{12},-\dfrac{5}{6} \right)$$ is the vertex of the given parabola and length of the latus rectum is $$\dfrac{1}{3}$$ . So the equation hence formed is the equation of the parabola in vertex form. **Note:** Any problem of conic section is very easy to solve once we know how to approach the problem and we keep in mind the different formulae and relations. To convert any equation of the parabola in vertex form we first of all need to rearrange the equation to form a perfect square. Now once we have done this, it becomes very easy for us to determine the value of the vertex as well as the latus rectum of the parabola.