parabola, ln, xyz points

[xcrush]

I just have a few questions I still can't figure out..

#1. Find the equation that describes the set of points (x, y, z) that are 5 units from the point (2, -1, 6).

I did 2x - 5 - y - 5 + 6z - 5 = 0 and got 2x - y + 6z = 15. But I'm almost positive that is wrong.

#2. Find an equation for the parabola whose vertex is (2, -5) and passes through (4, 7). Express your answer in the standard form for a quadratic function.

I have absolutely no idea how to do this one at all.

#3. Solve for x: e^2x = 3x^2.

All I got so far was x = ln3x^2 - ln2

#4. Find the points of intersection of x^2 + y^2 = 4 and x^2 + y^2 - 4x - 4y = -4, graphically and algebraically.

I know they are both circles, but I'm not sure how to find their points of intersection.

 

[pka]

Show us what you have tried on each one of these.

 

[stapel]

1) The three-dimensional Distance Formula might be a good start.

2) Use the vertex formula for a parabola. Plug the vertex in. Plug the x,y-coordinates of the point in. Solve for whatever remains.

3) How did you arrive at "x = ln(3x²) - ln(2)"? Please show your steps.

4) Using the "subtraction" method would probably be useful for solving this system, I think.

Eliz.

 

[soroban]

Hello, xcrush!

Here's #4 . . .

4) Find the points of intersection of \(\displaystyle \,\begin{array}{cc}x^2\,+\,y^2\:=\:4 \\x^2\,+\,y^2\,-\,4x\,-\,4y\:=\:-4\end{array}\)
graphically and algebraically.

We have: \(\displaystyle \,[1]\;x^2\,+\,y^2\:=\:4\)
\(\displaystyle \;\;\)and: \(\displaystyle \,[2]\;x^2\,+\,y^2\,-\,4x\,-\,4y\:=\:-4\)

Subtract [2] from [1]: \(\displaystyle \:4x\,+\,4y\:=\:8\;\;\Rightarrow\;\;y\:=\:2\,-\,x\;\;[3]\)

Substitute [3] into [1]:\(\displaystyle \:x^2\,+\,(2\,-\,x)^2\:=\:4\;\;\Rightarrow\;\;2x^2\,-\,4x\:=\:0\)

Factor: \(\displaystyle \,2x(x\,-\,2)\:=\:0\;\;\Rightarrow\;\;x\:=\;0,\;2\)

Substitute into [3]: \(\displaystyle \:y\:=\:2,\;0\)

Intersections: \(\displaystyle \2,0),\;(0,2)\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Equation [1] is a circle with center (0,0) and radius 2.

Equation [2] is: \(\displaystyle \,(x\,-\,2)^2 + (y\,-\,2)^2\:=\:4\)
\(\displaystyle \;\;\) a circle with center (2,2) and radius 2.

               |      * * *
               |   *         *
               | *             *
               |*               *
               |
               *      (2,2)      *
             *2◊ *      +        *
          *    *    *            *
        *      |      *
       *       |*      *        *
               | *             *
      *        |   *    *    *
   ---*--------+------*-◊-*----------
      *        |        *2
               |
       *       |       *
        *      |      *
          *    |    *
             * * *
               |

 

[xcrush]

thanks soroban!

here's what I did for #3..

e^2x = 3x^2

I took the natural log of each side and got:
2x = ln3x^2

then divide each side by 2:
x = ln(3x^2)/2

then convert the division into subtraction:
x = ln3x^2 - ln2


And for #2, I'm not sure what to do with the (4,7) part. My book explains it in a complicated way..

 

[daon]

here's what I did for #3..

e^2x = 3x^2

I took the natural log of each side and got:
2x = ln3x^2

then divide each side by 2:
x = ln(3x^2)/2

then convert the division into subtraction:
x = ln3x^2 - ln2

You are confusing the division/subtraction rule. log(a/b) = loga - logb. You have log(a)/b. Two different things.

And for #2, I'm not sure what to do with the (4,7) part. My book explains it in a complicated way..

For #2, you have P(4,7) and v=(2,-5)

Use the general formula: y = ax^2 + bx + c

You know the vertex is (2,-5), and that the x value, 2, is equal to -b/2a or b=-4a. You also know that 4,7 and 2,-5 are on the parabola. So, plug them into the formula. Now you know all the following:

1. b = -4a
2. 7 = a(4)^2 + b(4) + c
3. -5 = a(2)^2 + b(2) + c

You now have 3 equations to solve for three variables.