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 $y = -\frac{517}{475}\approx -1.088$.

 

[CellCoree]

The vertex of a parabola is the point where the parabola changes direction, also known as the highest or lowest point on the curve. In order to determine the coordinates of the vertex, we can use the formula $x_v = -\frac{D}{2A}$ and $y_v = -\frac{E}{2C}$, where $x_v$ and $y_v$ represent the x-coordinate and y-coordinate of the vertex, respectively.

In the given example, the general equation of the parabola is $4x^2 + 4xy + y^2 - 5x + 7y + 11 =0$. By comparing this with the general equation given in the problem, we can see that $A=4$, $B=4$, $C=1$, $D=-5$, and $E=7$. Substituting these values into the formula, we get $x_v = -\frac{-5}{2(4)} = \frac{5}{8}$ and $y_v = -\frac{7}{2(1)} = -\frac{7}{2}$. Therefore, the vertex of the parabola is $(\frac{5}{8}, -\frac{7}{2})$.