Image of a line or conic on Veronese surface. This is part of Exercise 5.13 from Undergraduate Algebraic Geometry by Reid:

Consider the Veronese surface $S$ defined by the map:
  $$\phi: \mathbb{P}^2\rightarrow \mathbb{P}^5$$
  where $\phi(x_0,x_1,x_2)=(x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2)$. 

The problem asks to show that a line in $\mathbb{P}^2$ is mapped to a conic in $\mathbb{P^5}$ and a conic in $\mathbb{P}^2$ is mapped to a quartic in $\mathbb{P}^5$. 
My attempt:
Suppose a line in $\mathbb{P}^2$ is defined by $ax_0+bx_1+cx_2=0$. Then we have also $V: ay_0+by_1+cy_2=0$ in $\mathbb{P}^5$. So the image of the line in $\mathbb{P}^5$ is the intersection of $V$ and $S$. I have the following questions:


*

*How do we show that it is a conic?

*How do we decide we don't need more equations? For example, $ay_1+by_3+cy_5=0$ can also define it. So does $ay_2+by_4+cy_5=0$. 


I read some pages by Harris. It has some nice description of the Veronese surface, but my questions are not solved. I have similar questions then about a conic mapped to quartic. 
Thank you for your help!
Edit:
Let $x_0^2+x_1^2+x_2^2=0$ be a conic in $\mathbb{P}^2$. Its image in $\mathbb{P}^5$ is the intersection of $y_0+y_3+y_5=0$ and the surface $S$. Making a change of variable so $y_5=0$ and plugging $-y_0-y_3$ into $y_5$ of the three defining equations of $S$, I got
$$y_1^2=-(y_3^2+y_4^2)\\
y_1^2=-(y_0^2-y_2^2)\\
y_1^2=y_0y_3$$
The pullback of the first two are union of the conic and a line ($x_0=0$ and $x_1=0$, respectively). How to write it as a single quartic so that the pullback does not contain the extra line?
 A: I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in $\mathbb{P}^{5}$  by $[z_{0}:\ldots:z_{5}]$ and the coordinates in $\mathbb{P}^{2}$ by $[x_{0}:x_{1}: x_{2}]$.
Claim The Veronese surface is cut out by three quadrics:
$$C_{1}:  z_{0}z_{3} - z_{1}^{2} = 0$$
$$C_{2}: z_{0}z_{5} -z_{2}^{2}= 0 $$
$$C_{3}: z_{3}z_{5} - z_{4}^{2} = 0. $$
It should be checked that they are indeed enough to define the locus but we will not get into that.
We consider a quadric $Q$ in $\mathbb{P}^{2}$ given by $Q: a_{0}x^{2}_{0} -(a_{1}x^{2}_{1} + a_{2}x^{2}_{2}) = 0 $. We can and will assume that $a_{0} \neq 0$ and after rescaling $a_{0} = 1$.
Clearly the image of $Q$ is contained in the hyperplane $H: z_{0} - a_{1}z_{3} - a_{2}z_{5} = 0$.
Our goal: Understand the intersection of this hyperplane with quadrics $C_{i}$.
$$C_{1} \cap H : (a_{1}z_{3} + a_{2}z_{5})z_{3}-z_{1}^{2}= 0$$
$$C_{2} \cap H : (a_{1}z_{3} + a_{2}z_{5})z_{5}-z_{2}^{2}= 0$$
Multiplying the first equation above by $z_{5}$ and the second one by $z_{3}$ and subtracting them we get the equation $E:z_{1}^{2}z_{5} - z_{2}^{2}z_{3} = 0.$
This equation $E$ represents the locus of the intersection of $H$ with two of the three quadrics. The justification for this ad-hoc process is the the following:
Inside the open (quasi-affine) variety $z_{3}z_{5} \neq 0$ we are allowed to multiply by non-zero  functions $z_{3}$ and $z_{5}$. Outside this locus one has to check that this equation is still valid.
Next we study the intersection of $E$ and $C_{3}$. Again one multiplies $E$ by $z_{3}$ and $C_{3}$ by $z_{1}^{2}$ subtract the resulting equations and get $z_{1}^{2}z_{4}^{2}- z_{2}^{2}z_{3}^{2} = 0$ - a quartic.
The computation of the image of a cubic in $\mathbb{P}^{2}$ under the Veronese in Harris's book, page 2  here , is a great illustration of the complexity of figuring out the intersections.
