# Parabola problem

• Feb 20th 2011, 04:49 PM
frankinaround
Parabola problem
Hi. heres the problem :
1. Find the equation of the parabola with focus (1,1) and directrix x+y=0, and simplify this equation to a form without radicals.

2.Let the vertex of the parabola x^2=4py be joined to every other point of the parabola. show that the midpoints of the resulting chords lie on another parabola. find the focus and directrix of the second parabola.

for the first one, I dont get how the to get the directrix. if its x+y=0 and the focus is (1,1), then the directrix should be 1-2(1/4A) or 1/2A. (because the focus is (x,y1-1/4A) so 1-1/4A should be y1 and 1-1/4A-1/4A should be the directrix. (I think). im confused as to where to go though. Any tips on this one?

the second one : im confused about the letter P. Maybe 4p = A, and the equation if I move it around is :
1/4p(x-0)^2=(y-0)
if this is true then the vertex is (0,0). But Im kind of confused. I cant find exact points because I have this variable p and I dont get it. but I kind of think that like... this equation : 1/4p(x/2)^2=y/2 might be the equation of the second parabola. not sure about that though. Cant find the focus and directrix of first parabola because I dont get p. Please help.
• Feb 20th 2011, 11:17 PM
earboth
Quote:

Originally Posted by frankinaround
Hi. heres the problem :
1. Find the equation of the parabola with focus (1,1) and directrix x+y=0, and simplify this equation to a form without radicals.

...

for the first one, I dont get how the to get the directrix. if its x+y=0 and the focus is (1,1), then the directrix should be 1-2(1/4A) or 1/2A. (because the focus is (x,y1-1/4A) so 1-1/4A should be y1 and 1-1/4A-1/4A should be the directrix. (I think). im confused as to where to go though. Any tips on this one?

1. Use the definition of a parabola as the locus of points whose distance from a fixed point (focus) equals the distance to a fixed straight line (directrix).

2. Let P(x, y) denote the point on the parabola. Then the distance to the line is:

$d_d = \dfrac{1 \cdot x + 1 \cdot y}{\sqrt{1^2+1^2}} = \dfrac{x+y}{\sqrt{2}}$

3. The distance of P to F(1, 1) is

$d_F=\sqrt{(x-1)^2+(y-1)^2}$

4. Both distances are equal. Re-aarange the equation until it doesn't contain any roots.
Spoiler:
You should come out with $\frac12(x-2)^2+\frac12(y-2)^2-xy=2$
• Feb 20th 2011, 11:37 PM
earboth
Quote:

Originally Posted by frankinaround
Hi. heres the problem :
...

2.Let the vertex of the parabola x^2=4py be joined to every other point of the parabola. show that the midpoints of the resulting chords lie on another parabola. find the focus and directrix of the second parabola.

...

the second one : im confused about the letter P. Maybe 4p = A, and the equation if I move it around is :
1/4p(x-0)^2=(y-0)
if this is true then the vertex is (0,0). But Im kind of confused. I cant find exact points because I have this variable p and I dont get it. but I kind of think that like... this equation : 1/4p(x/2)^2=y/2 might be the equation of the second parabola. not sure about that though. Cant find the focus and directrix of first parabola because I dont get p. Please help.

1. The general equation of a parabola is $(x-k)^2=4p(y-h)$ where p is the distance between the vertex and the focus.

2. The vertex has in your case the coordinates V(0, 0). Let $P (k, \frac1{4p} k^2)$ a point on the given parabola. The the midpoint of the chord $\overline{VP}$ is $M\left(\frac12 k,\ \frac1{8p} k^2\right)$.

3. All points M form a curve. You know that the x-value is: $x = \frac12 k ~\implies~k = 2x$
You know that the y-value is: $y = \frac1{8p} k^2$
Plug in the term for k:
$y = \frac1{8p}(2x)^2 = \frac1{2p}x^2$

4. The equation of the new parabola is: $2py = x^2$
Now determine the vertex and the focus of the new parabola and compare them with the corresponding values of the given parabola.
• Feb 22nd 2011, 06:56 PM
frankinaround
Quote:

Originally Posted by earboth
2. Let P(x, y) denote the point on the parabola. Then the distance to the line is:

$d_d = \dfrac{1 \cdot x + 1 \cdot y}{\sqrt{1^2+1^2}} = \dfrac{x+y}{\sqrt{2}}$

I understand the distance from the point to the focus, but I dont get how you came up with the point to the directrix. Can you help out with that part?
• Feb 22nd 2011, 07:06 PM
frankinaround
Quote:

Originally Posted by earboth
1. The general equation of a parabola is $(x-k)^2=4p(y-h)$ where p is the distance between the vertex and the focus.

2. The vertex has in your case the coordinates V(0, 0). Let $P (k, \frac1{4p} k^2)$ a point on the given parabola. The the midpoint of the chord $\overline{VP}$ is $M\left(\frac12 k,\ \frac1{8p} k^2\right)$.

3. All points M form a curve. You know that the x-value is: $x = \frac12 k ~\implies~k = 2x$
You know that the y-value is: $y = \frac1{8p} k^2$
Plug in the term for k:
$y = \frac1{8p}(2x)^2 = \frac1{2p}x^2$

4. The equation of the new parabola is: $2py = x^2$
Now determine the vertex and the focus of the new parabola and compare them with the corresponding values of the given parabola.

equation of a parabola = A(x-x1)^2=(y-y1)
focus = (x1, y1+(1/4a))
directrix = (y1-(1/4a))
vertex = (x1,y1)

thats why the p confused me. So I would like to know is
A) is the method I learned correct? if so whats the difference between 4P and A ?
and B) why did you choose $P (k, \frac1{4p} k^2)$ as a point on the parabola?
• Feb 23rd 2011, 12:23 AM
earboth
Quote:

Originally Posted by frankinaround
I understand the distance from the point to the focus, but I dont get how you came up with the point to the directrix. Can you help out with that part?

I don't know how much you know about vector geometry (then this problem could be explained very easily). So the best for you would be to memorize a formula:

The distance between a point Q(m, n) and the straight line $l: Ax+By=C$ is calculated by:

$d(Q,l)=\dfrac{Am + Bn - C}{\sqrt{A^2+B^2}}$

If you compare this general form of the formula and my calculations you'll see that I've strictly used the formula.
• Feb 23rd 2011, 12:34 AM
earboth
Quote:

Originally Posted by frankinaround
Maybe I was taught wrong or something. No
equation of a parabola = A(x-x1)^2=(y-y1)
focus = (x1, y1+(1/4a))
directrix = (y1-(1/4a))
vertex = (x1,y1)

thats why the p confused me. So I would like to know is
A) is the method I learned correct? if so whats the difference between 4P and A ?
and B) why did you choose $P (k, \frac1{4p} k^2)$ as a point on the parabola?

Your method and the equation I prefer are exactly the same:

$p:(x-x_1)^2=4p(y-y_1)$ where p = distance between the vertex and the focus. You have learned that the distance between the vertex and the focus is $\dfrac1{4A}$. Now I plug in your value into my equation:

$p:(x-x_1)^2=4 \cdot \dfrac1{4A}(y-y_1)$

$p:(x-x_1)^2= \dfrac1{A}(y-y_1)$

$p:A(x-x_1)^2= (y-y_1)$

In my opinion my equation is a little bit more comfortable because you can see at once the distance between the vertex and the focus.

To your 2nd question: I wanted to get an equation in x and y. So I had to use a different name of the coordinates of P to avoid the situation that I have an equation where "x" has two different meanings.
• Feb 24th 2011, 08:08 PM
frankinaround
Quote:

Originally Posted by earboth
1. The general equation of a parabola is $(x-k)^2=4p(y-h)$ where p is the distance between the vertex and the focus.

2. The vertex has in your case the coordinates V(0, 0). Let $P (k, \frac1{4p} k^2)$ a point on the given parabola. The the midpoint of the chord $\overline{VP}$ is $M\left(\frac12 k,\ \frac1{8p} k^2\right)$.

3. All points M form a curve. You know that the x-value is: $x = \frac12 k ~\implies~k = 2x$
You know that the y-value is: $y = \frac1{8p} k^2$
Plug in the term for k:
$y = \frac1{8p}(2x)^2 = \frac1{2p}x^2$

4. The equation of the new parabola is: $2py = x^2$
Now determine the vertex and the focus of the new parabola and compare them with the corresponding values of the given parabola.

can you tell me why in this random point on parabola you chose $P (k, \frac1{4p} k^2)$ 1/4p * k^2 as y? Why didnt you just say x,y is a point on the parabola?
• Feb 24th 2011, 10:46 PM
earboth
Quote:

Originally Posted by frankinaround
can you tell me why in this random point on parabola you chose $P (k, \frac1{4p} k^2)$ 1/4p * k^2 as y? Why didnt you just say x,y is a point on the parabola?

OK, I'll use your suggestion ...

1. The given parabola has the equation $x^2=4py$. Therefore the vertex is V(0, 0).

2. Let P(x, y) denote a point on the parabola. Then the midpoint of $\overline{VP}$ is $M_{\overline{VP}}\left(\frac12 x,\ \frac12 y \right)$

3. All points M form a curve. You know that the x-values are:

$x_M = \frac12 x$
and the y-values are:
$y_M = \frac12 y$

4. And now ... it's your turn! You must now use the equation of the parabola and after few additional steps you'll get an equation in $x_M$ and $y_M$ which I reached in a single step by using the variable k as abscissa of point P.
• Feb 26th 2011, 05:44 PM
frankinaround
I mean, I dont want to look stupid... but I have no choice.

So I already figured out that the midpoint would be x/2,y/2 (from midpoint of new parabola to origin). But the reason I asked why you used (K, k^2/4p) is because I couldnt figure out how to derive an equation for parabola having x/2 and y/2.

i mean, the parabola should be the distance between the point on the new parabola and the directrix and the point and the focus :

sqrt of (x/2)^2+(y/2)^2 = ((-1/4a)y/2) / (1/4a)

sqrt (x/2)^2+(y/2)^2 becomes x^2/4 + (y^2/4 ) - (y/8a) - (y/8a) + (1/16a^2) so

sqrt(x^2/4 + (4y^2A^2-4ya+1)/16a^2) = -y/2

I tried to do binomial theorum on : 4y^2^2-4ya+1 to maybe simplify it but im getting a - under a sqrt.
also I can try squaring both sides and bringing over the right and then but binomial theorum would still give me anegative under radical for that.

so yeah.. thats why I was wondering why you used that variable instead of x,y haha.... ugh. (because I keep getting stuck with this problem)
• Feb 27th 2011, 01:24 AM
earboth
Quote:

Originally Posted by frankinaround
I mean, I dont want to look stupid... but I have no choice.

So I already figured out that the midpoint would be x/2,y/2 (from midpoint of new parabola to origin). But the reason I asked why you used (K, k^2/4p) is because I couldnt figure out how to derive an equation for parabola having x/2 and y/2.

i mean, the parabola should be the distance between the point on the new parabola and the directrix and the point and the focus :

sqrt of (x/2)^2+(y/2)^2 = ((-1/4a)y/2) / (1/4a)

...

When you get a new parabola you also get a new focus and a new directrix. As far as I understand your calculations you used the directrix and the focus of the original parabola. By this method you'll never get a correct result.

I used the midpoint of the chord as an element of a locus of points. The coordinates of an element of a locus of points depend on a parameter which I labeled k. Then you have to eliminate the parameter to get the equation of the locus. (see post #3)

I've attached a construction on how you get the new parabola from the original parabola by constructing the points M.