How to Determine the Vertex of a Parabola?

[Amer]

The general equation of the parabola has the form
$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ with $B^2 = 4AC$
what is the coordinate of the vertex of the parabola or in other word how to determine the vertex of a given parabola for example
$4x^2 + 4xy + y^2 - 5x + 7y + 11 =0 $

Thanks.

 

[mathmari]

The general equation of the parabola has the form
$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ with $B^2 = 4AC$
what is the coordinate of the vertex of the parabola or in other word how to determine the vertex of a given parabola for example
$4x^2 + 4xy + y^2 - 5x + 7y + 11 =0 $

Thanks.

The general equation of the parabola has the form
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, \text{ with } A \neq 0, C=0$$
OR
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, \text{ with } A=0, C \neq 0$$

$4x^2 + 4xy + y^2 - 5x + 7y + 11 =0$ is an ellipse.

 

[Opalg]

The general equation of the parabola has the form
$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ with $B^2 = 4AC$
what is the coordinate of the vertex of the parabola or in other word how to determine the vertex of a given parabola for example
$4x^2 + 4xy + y^2 - 5x + 7y + 11 =0 $

Thanks.

In general, the axis of the parabola will be at an angle to the coordinate axes. I think that the easiest way to find the vertex is probably to find some new coordinates $(X,Y)$ so that the axis of the parabola is parallel to one of the new coordinate axes.

If the equation is $4x^2 + 4xy + y^2 - 5x + 7y + 11 =0 $, then you can write it as $(2x+y)^2 = 5x - 7y - 11$. Take one of the new coordinates to be $Y = 2x+y$. The other coordinate $X$ must be perpendicular to $Y$, so take it to be $X = x-2y$. Next, solve the equations to get the old coordinates $(x,y)$ in terms of the new ones. That gives $x = \frac15(X+2Y)$, $y = \frac15(Y-2X)$. Substitute those values into the parabola equation, getting $Y^2 = (X+2Y) - \frac75(Y-2X) + 11$.

At this stage, the arithmetic gets messy, so I won't try to get the numbers correct. But in principle you can rearrange that last equation to get it into the form $(Y-\alpha)^2 = \beta(X-\gamma)$ for some coefficients $\alpha$, $\beta$, $\gamma$. You should be able to recognise that as a parabola with vertex at $(X,Y) = (\gamma,\alpha)$. Now all that remains is to express that in terms of the old coordinates $(x,y)$.

 

[Evgeny.Makarov]

The general equation of the parabola has the form
$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ with $B^2 = 4AC$

Strictly speaking, it is also required that
\[
\begin{vmatrix}
A & B/2 & D/2\\
B/2 & C & E/2\\
D/2 & E/2 & F
\end{vmatrix}\ne0.
\]

The general equation of the parabola has the form
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, \text{ with } A \neq 0, C=0$$
OR
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, \text{ with } A=0, C \neq 0$$

If $A=0$ or $C=0$ while $B\ne0$, then it is not a parabola since $B^2\ne4AC$.

$4x^2 + 4xy + y^2 - 5x + 7y + 11 =0$ is an ellipse.

No, it's a parabola.

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The vertex of a parabola with equation $F(x,y)=0$ can be found as follows. First find the asymptotic direction $(\alpha,\beta)$ from the equation
\[
A\alpha^2+B\alpha\beta+C\beta^2=0
\]
(there are infinitely many nonzero solutions). Then the equation of the parabola axis is
\[
-\beta\frac{\partial F}{\partial x}+\alpha\frac{\partial F}{\partial y}=0\qquad(*)
\]
so the coordinates of the vertex can be found from (*) and $F(x,y)=0$.

In the case of
\[
4x^2 + 4xy + y^2 - 5x + 7y + 11 =0
\]
the asymptotic direction is $(1,-2)$, the axis is $20x+10y-3=0$ and the vertex is $x = \frac{1319}{1900}\approx 0.694$ and $y = -\frac{517}{475}\approx -1.088$.