Actually, I got the idea of solving it from the chat room via @BAYMAX . The original equation in general form is as follows :


In order to find the vertex, I take $x+y=p$ . Then,

$$p^2 +6(p-y)+4y+3=0$$ $$or, p^2+6p-2y+3=0$$ $$or, (p+3)^2 = 4\times \frac{1}{2} (y+3)$$

The equation is now in $p-y$ plane. For finding vertex, I get $p=-3 , y=-3 $ ... which renders $x=0$ .

This, the vertex is (0,-3) . Am I correct? Besides, how do I get the focus?

  • $\begingroup$ Consider removing the anecdotic comments. $\endgroup$ – Yves Daoust Aug 25 '18 at 11:48

The standard way is that of switching to a pair of rotated axes: $X=(x-y)/\sqrt2$, $Y=(x+y)/\sqrt2$. You'll get then a standard parabola equation in the $(X,Y)$ plane and once you've found the vertex you can switch back to $(x,y)$. The result for the vertex turns out to be $(3/8,-23/8)$. Your method is flawed because your coordinate transformation is not an isometry or a similitude.

If you don't want to use rotations, I can show you some alternate method.


First of all we must find the direction of the axis. Notice that every line parallel to the axis intersects the parabola at a single point, i.e. substituting the equation of the line into the equation of the parabola must give a linear equation. This is possible only if the quadratic part of the parabola equation is a constant, that is: $x+y=a$. This is then the equation of a generic line parallel to the axis, and it is bent at an angle of 45° with the $y$ axis; to have a standard parabola equation we must then perform a 45° counterclockwise rotation of the axes.

But we can find the vertex even without rotations: a generic line perpendicular to the axis has equation $y=x+b$ and the vertex is the only point where such a line meets the parabola at a single point. Substituting the equation of the line into that of the parabola gives: $$ (2x+b)^2+10x+4b+3=0, \quad\text{that is:}\quad 4x^2+(4b+10)x+b^2+4b+3=0. $$ This has a single solution if its discriminant vanishes, that is if: $$ (4b+10)^2-16(b^2+4b+3)=0, \quad\text{whence:}\quad b=-{13\over4}. $$ With this value of $b$ the solution of the intersection equation is $$ x={-(4b+10)\over8}={3\over8}, \quad\text{and}\quad y=x+b=-{23\over8}. $$ These are then the coordinates of the vertex.


There is another way, if you want, to find the direction of the axis, exploiting this property of any parabola: the line joining the midpoints of two parallel chords is parallel to the axis.

Take then two parallel lines intersecting your parabola, e.g.: $$ y=0,\ \text{intersecting at:}\ x=-3\pm\sqrt{6},\ \text{midpoint:}\ M_1=(-3,0)\\ y=1,\ \text{intersecting at:}\ x=-4\pm\sqrt{8},\ \text{midpoint:}\ M_2=(-4,1). $$ Line $M_1M_2$ has equation $y+x=-3$ and is then parallel to the axis.

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  • $\begingroup$ Could you please tell me how you got the equation for rotation.... And is there other method? I would love to learn them $\endgroup$ – Entrepreneur Aug 26 '18 at 6:09
  • $\begingroup$ I edited my answer. $\endgroup$ – Intelligenti pauca Aug 26 '18 at 10:58
  • $\begingroup$ You've used a line perpendicular to the axis in second method and substituted to get the answer. Why can't we use $x+y=a$ here? $\endgroup$ – Entrepreneur Aug 26 '18 at 12:36
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    $\begingroup$ The two cases are different: - ANY line $y=kx+a$ parallel to the axis must have a single intersection, hence the resolvent equation has a single solution FOR EVERY VALUE of parameter $a$; the only way to get this is that of obtaining a linear equation for any $a$. - A line $y=x+b$ perpendicular to the axis can have two, one or no intersection, depending on the value of $b$; the resolvent equation is quadratic, but its discriminant vanishes for a single value of $b$, the one corresponding to the vertex. $\endgroup$ – Intelligenti pauca Aug 26 '18 at 14:16

The vertex is the intersection of the parabola with its axis of symmetry; see this.

The resulting system $$\langle (x+y)^2+6x+4y+3,x+y+\frac52\rangle$$ has solution $$(x,y)=(\frac38,-\frac{23}{8}).$$


For the focus I only know of the method by rotating i.e. substitute $x=\frac{x'+y'}{\sqrt{2}}, y=\frac{y'-x'}{\sqrt{2}}$ into the equation to get:


i.e. $x''=-\frac1{4\frac1{4\sqrt{2}}}y''^2$ which has focus $(x'',y'')=(-\frac1{4\sqrt{2}},0)$ (cf. wikipedia). So the intermediary equation has focus $(x',y')=(3/\sqrt{2},-5/(2\sqrt{2}))$ which transforms back to $(x,y)=(\frac14,-\frac{11}{4}).$

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  • $\begingroup$ Is there any technique for finding the focus ? $\endgroup$ – Entrepreneur Aug 26 '18 at 10:00
  • $\begingroup$ @Entrepreneur: I edited my answer. $\endgroup$ – Jan-Magnus Økland Aug 26 '18 at 11:08

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