$(x+2y)^2=4(x-y)$
Disecting the above parabola is the question. (vertex, axis,tangent at vertex,etc).
So at first what I thought of was making its equations at LHS and RHS perpendicular. I thought since it might be tedious (which wasnt at all and got me the correct answer), I thought of something else. I thought that a parabola is symmetric bout its axis. Consider $y^2=4ax$. I get two values of y for every +ve x and that is why its symmetric. But what if I put something in x such that y has no option except one i.e. the terms containing y make a perfect square. So I expanded the above and wrote it in form of a quadratic of second degree in y and put $b^2=4ac$. Ofcourse, a,b and c were in terms of x, which yielded a value which I thought was the answer but the answer came by introducing a k in the square term in LHS and making slope 2 in RHS. But what is wrong in my method?
[Math] wrong with this method for a rotated and shifted parabola
conic sectionscoordinate systems
Related Solutions
As far as your diagram looks, you might try some ombination of parabolas to fit your function..
Your lower part (both sides) look like more or less from a same parabola, andthe upper part from a different one. So you might make a function like $f(x)$=Parabola A when y coordinates below a certain point and Parabola B otherwise.
Now what these two (or more) parabolas are, you yourself have to find out , by brute force or something..
For convenience I assume x-axis as directrix. Variable focus is at $(h,k)$ which should be eliminated by differentiation.
$$ (x-h)^2 + (y-k)^2 =y^2 \tag1 $$ $$ x^2 -2 x h -2 y k + h^2+k^2 =0 \tag2$$ Differentiating with respect to x $$ y^{\prime } k = x-h \tag3$$ Differentiating with respect to x again $$ y^{\prime \prime} =1/k\tag4 $$ Solving for $(h,k)$ for elimination
$$(h,k) = ( x- \frac{ y^{\prime }}{y^{\prime \prime }},\,\frac{ {1 }}{y^{\prime \prime }}) \tag5$$
Plug this into (1) and simplify to get the required second order differential equation of arbitrary focus location. The ODE with x-axis as directrix and axis of symmetry not necessarily coinciding with y-axis is given by:
$$ 2 y y^{\prime\prime} = 1+ y^ {{ \prime}^2} \tag6$$
So now recognizing this curvature ratio property ( may be mentioned in old (Chelsea Publishers?) Conics text-books).. The red curve is the parabola evolute.
The radius of curvature of a parabola is double the normal segment length intercepted by the directrix.
$$ \dfrac{ y y^{\prime\prime}} { 1+ y^ {{ \prime}^2}} = \frac{R_d}{R_c}=\frac12\tag7$$
The offset axis case has been numerically computed and verified by CAS It is shown ( BC to arbitrary start point and slope) at right.
So this property can be used beneficially to define parabolic arc differential equation.
Change of axes gives the ODE of parabola whose axis is parallel to x-axis and displaced arbitrarily parallel with respect to $x-$ axis. The focus in general is not on x-axis.
$$ 2 x y^{\prime\prime} = -y^{\prime}(1+y^{{ \prime}^2}) \tag8$$
This is not preferred ODE in the unsymmetric case as there are two distinct values of $y$ for same $x$, it is a mess.
EDIT1:
In hindsight perhaps the easiest way is to get ODEs of all parabolas on same x-axis directrix is Formula17_MW_WA. Curvature ratio
$$ \frac{PD}{PC}= = \frac{R_d}{R_c}=\dfrac{(a t^2+a) \sec\phi }{2a \sec^{3}\phi} =\frac12=\dfrac{ y y^{\prime\prime}} { 1+ y^ {{ \prime}^2}}. \tag9$$
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Best Answer
This parabola cannot be written in $x=f(y)$ form (like a sideways parabola) or in $y=f(x)$ form (vertical parabola). You can see this by the mixed $xy$ term when you expand everything. It is a parabola rotated to some arbitrary angle. That's why none of your methods work. You need to first find out the rotation angle, then use a coordinate transformation to put it in a standard form and then you can do all the good things you want to do with it.
If your parabola is of the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ then the angle by which the parabola has been rotated is given by $$\cot(2\theta)=\frac{A-C}{B}.$$
In your case you have $x^2+4xy+4y^2-4x+4y=0$ so $$\theta = \frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\approx-26.56^{\circ}\approx333.43^{\circ}$$
and indeed your parabola looks like
The thing here is that we actually don't need $\theta$ itself but rather we need $\sin(\theta)$ and $\cos(\theta)$ to "undo" the rotation. We can get the exact values without computing $\theta$ itself by drawing a triangle and using half-angle identities
$$\cos(\theta)=\cos\left(\frac{1}{2}\arctan(-4/3)\right)=\frac{2}{\sqrt{5}}$$ $$\sin(\theta)=\sin\left(\frac{1}{2}\arctan(-4/3)\right)=-\frac{1}{\sqrt{5}}$$
and then substitute (call this set 1)
$$x = x'\cos(\theta)-y'\sin(\theta)$$ $$y = x'\sin(\theta)+y'\cos(\theta)$$
and simplify the given parabola to get
$$-12x'+4y'+5\sqrt{5}y'^2=0.$$
So now we have a sideways parabola and notice how the mixed term $xy$ has disappeared so now you get do whatever you need to do to this rotated parabola. But after you find the directrix, vertix, focus, etc, you would have to rotate all those by $\theta$ above to get the correct values in the correct coordinate system. So we unrotate the parabola, find everything we need, and then rotate everything we found.
Continuing this example, the sideways parabola looks like this.
It is very slightly shifted and scaled. I leave the rest of algebra up to you. After you find everything you need, for some of them you will need (call this set 2)
$$x' = x\cos(\theta)+y\sin(\theta)$$ $$y' = -x\sin(\theta)+y\cos(\theta)$$
to go back to the original coordinate system.
To go from $(x',y')\rightarrow (x,y)$ use set 1. To go from $(x,y)\rightarrow (x',y')$ use set 2.
Addendum: Your method doesn't work because it only works with unrotated (canonical) parabolas meaning parabolas where the axis of symmetry is parallel to the $x$-axis or the $y$-axis. The methods you are using don't work when there is a mixed $xy$ term present so we change coordinates and eliminate the $xy$ term and then use the "shortcuts" you know. Like Blue said, you can use the easy (shortcut) methods in a complicated way or you can use complicated methods (basics/using definitions) in a simple way. For example, the definition of a parabola is that given a point (the focus) and a straight line (a directrix), a parabola is the set of all points equidistant from both. For any parabola rotated or not, you can use this definition to find the focus and the directrix. But if you want to use the
"Start with $y=ax^2+bx+c$ and put it into $4p(y-k)=(x-h)^2$ and your focus stares at you in your face with coordinates $(h,k)$ and $p$ is the distance from the vertex to the focus, and the directrix"
then your parabola MUST be a canonical form (in this case opening up or down). This shortcut for finding the focus, vertex, and directrix ONLY works if there is no $xy$ term. And that is because this shortcut is derived assuming that the parabola opens only up or down. It cannot be sideways nor at any other angle. Since your parabola isn't canonical, these shortcuts fail. So you either "make it canonical" or you use the more complicated/definition method like Blue's method. They both work. You decide which one you prefer.