Answer to Question #114936 in Analytic Geometry for ANJU JAYACHANDRAN

Question

Reduce the following equations to standard form, and then identify which conicoids they represent. Further, give a rough sketch of each.
i) x^2 +y^2 +2x−y−z+3 = 0
ii) 3y^2 +3z^2 +4x+3y+z = 9

Answer 1

We know that for any given conicoids if we translate (shift) the origin (axes) suitably, $O$ to new origin (Axes) $O'$ ,then the coefficients of the linear terms in the conicoids vanishes and here we will use this idea to reduce the coinicoids (i) and (ii).

i). We have given that the non standard form of the conicoid is

$$x^2 +y^2 +2x−y−z+3 = 0 \hspace{1cm} (1)$$

which looks like,

Let us consider such a suitable transformation of axes is,

$$x=x'-h\\ y=y'-k\\z=z'-l$$

Thus,Old origin $O=(0,0,0)$ and new origin $O'=(h,k,l)$.Now on evaluating the equation $(1)$ and equating the coefficients of the linear terms we get $(h,k,l)=(1,-\frac{1}{2},-\frac{7}{4})$ ,Hence the equation $(1)$ reduced to

$$(x+1)^2+(y-\frac{1}{2})^2=(z-\frac{7}{4})$$

Thus, standard form of the conicoid will be

$$x'^2+y'^2=z'$$

Which is a equation of paraboloid.

ii). Given conicoid is

$$3y^2 +3z^2 +4x+3y+z = 9\\ \implies y^2 +z^2 +\frac{4}{3}x+y+\frac{1}{3}z- 3=0 \hspace{1cm} (2)$$

Which looks like,

Now, exactly similar to the question (i) if we transform the axes,

$$x=x'-\frac{59}{24}\\ y=y'-\frac{1}{2}\\ z=z'-\frac{1}{6}$$

Thus, the standard form is

$$-\frac{3}{4}y'^2-\frac{3}{4}z'^2=x'$$

Which is also a paraboloid.