+118728 
+118728 Here is one you tube, I have not watched it but it will probably help.
There are many others if you don't like this one :)
https://www.youtube.com/watch?v=1Lwi8cOdqCU
Parabola
\(\boxed{(x-h)^2=4a(y-k)}\)
Vertex is (h,k)
a not equal 0
concave up if a>0 positive
concave down if a<0 negative
The focal length is |a|
The focal point will be (h,k+a)
The length of the latus rerctum is 4a The equation of it will be y= k+a
so
The points at the end of the latus r****m will be (h-2a,k+a) and (h+2a,k+a)
What else do you want to know?
Here is an excellent graphing calc so you can play around with some different equations and see what happens.
+118728 Hi Dimkorak,
Sorry I do not speak spanish but maybe you can make sense of this.
Maybe if you ask questions in Spanish someone else might be able to interpret for us.
Lo siento no hablo a español pero tal vez puede hacer sentido de esto.
Tal vez si preguntas en Español alguien podría ser capaz de interpretar para nosotros.
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Find the number in the interval [1/3, 2], such that the sum of the number and its reciprocal are a maximum.
Let the number be x
\(S=x+\frac{1}{x}\\ S=x+x^{-1}\\ \frac{dS}{dx}=1-1x^{-2}\\ \frac{dS}{dx}=1-\frac{1}{x^2}\\\)
Stat point when S'=0
\(0=1-\frac{1}{x^2}\\ \frac{1}{x^2}=1\\ x=\pm1\qquad but\quad 1/3\le x\le2 \qquad so\\ x=1 \)
\(\frac{d^2S}{dx^2}=2x^{-3}\\ \text{when x=1}\\ \frac{d^2S}{dx^2}=2>0\qquad so\\ \text{There is a minimum at x=1}\)
so the maximum must occur at one of the end points.
\(When \;\;x=1/3 \qquad S=3\frac{1}{3}\\ When\;\;x=2 \qquad \;\;\; S=2\frac{1}{2}\\~\\ \)
So the maximum will sum of the number and its reciprocal will be when the number is \(\frac{1}{3}\)
Here is a graph that you should try and make sense of.
https://www.desmos.com/calculator/zxw8w8frr9
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