## anonymous one year ago I need help! medal and fan

1. anonymous

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2. anonymous

2. Find the standard form of the equation of the parabola with a focus at (0, -10) and a directrix at y = 10. (5 points) y = x2 y = x2 y2 = -40x y2 = -10x 3. Find the standard form of the equation of the parabola with a focus at (5, 0) and a directrix at x = -5. (5 points) y = x2 20y = x2 x = y2 y2 = 20x

3. anonymous

@Thesn1

4. Owlcoffee

In analytical geometry, which is algebra applied to classic geometry, and I'll say this as a personal tip to you. Try interpreting what information you are given so you can narrow your options in order to get to the result much quicker (which is what problem-solving in mathematics is all about). I'll help you out on that. Let's begin with the first one you stated: "2. Find the standard form of the equation of the parabola with a focus at (0, -10) and a directrix at y = 10." At this point I'll suppose you know what a parabola is, as a geometric body, but I'll refresh that a little, we define "parabola" as the geometric place composed by all the points that are equally distant from a point called "focus" to a line called "directrix". That's pretty much it, no magic behind it. Now, let's analyze the given information, to narrow the options... We can observe that we are given the foci, and the directrix and these have the following conditions: - The foci is situated on the y-axis, because the x-coordinate is zero. -the directrix is a line, parallel to the x-axis, and cuts the y-axis in the point (0,10). This must mean that the parabola, follows the structure of: $y=\frac{ 1 }{ 2p }x^2$ but the foci is situated on the negative region of the y-axis, and that implies that the parabola opens downwards, so therefore: $y=-\frac{ 1 }{ 2p }x^2$ But, what is "p"?... Well, "p" is the distance between the foci and the directrix, and we can find it with a little of thought: $dist(F,V)=p$ "V" is the point defined by the intersection of the y-axis and the directrix, which is (0,10). now, the "vertex" of the parabola, is situated in the origin, because it is the mid-point between the foci and the directrix, or more specific, the mid-point between "V" and "F". $V \left[ (\frac{ 0+0 }{ 2 }),(\frac{ 10-10 }{ 2 }) \right]$ $V(0,0)$ So, therefore, if the mid point is the origin, then the distance must be the absollute value of the difference, you might know it from the distance in one dimension: $dist(F,V)=\left| 10-(-10) \right|$ $dist (F,V)=20$ but the distance from F to V was "p", so we can plug it in the equation, to find the values we were looking for: $y=-\frac{ 1 }{ 2(20) }x^2$ $y=-\frac{ 1 }{ 40 }x^2$ And that is the equation of the parabola defined by the given information.

5. Loser66

Another way: since the focus is (0,-10) , your p is -10, and the parabola is downward. Hence, the standard form of a downward parabola is $$x^2= 4py$$, replace all, you have $$x^2= 4(-10)y\\x^2= -40y\\y=-\dfrac{1}{40}x^2$$

6. Loser66

Same as problem 2, you have FORMULA for a parabola which is open to the right. Look up the formula from your note and find the solution out by yourself.