Tough Parabola

**c_323_h** · April 16th 2006, 05:26 PM

I found this to be a little tough.

Find the equation of a parabola with a vertex at the origin and focus at (1,1).

**ThePerfectHacker** · April 16th 2006, 06:35 PM

Originally Posted by c_323_h

I found this to be a little tough.

Find the equation of a parabola with a vertex at the origin and focus at (1,1).

It seems tough. I did not do it yet, do maybe this would help you. This is not your ordinary parabola i.e. sidewise or normal shaped. This one is a slanted parabola.

Since the vertex is at $(0,0)$ and focus $(1,1)$ then, drawing a line between them and extending it in the opposite direction the same distance you end up at $(-1,-1)$ . That means the directrix passes through this point. Furthermore, it is perpendicular to the line you drawn. Thus, its slope is $-1$ using slope and intercept-formula we have,
$y+1=-1(x+1)$
Thus,
$y=-x-2$ is equation of directrix.
Next, let $(x,y)$ be any point on the parabola. Then, you know that, the distance to the focus need to be equation to the distance from the line. Thus,
$\sqrt{(x-1)^2+(y-1)^2}=\frac{|x+y+2|}{\sqrt{2}}$
Do not worry the abosolute value disappers when you square both sides.
-------
I used the formula given a non-vertical line,
$Ax+By+C=0$ and a point not on the line, $(x_0,y_0)$ then the distance is,
$\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

Hope this helps.

**c_323_h** · April 16th 2006, 06:55 PM

yes, i found the equation of the directrix. the answer in my book is $x^2+y^2-8x-8y-2xy=0$

also, the distance between the origin and (1,1) is $\sqrt{2}$

**ThePerfectHacker** · April 17th 2006, 07:36 AM

Originally Posted by c_323_h

yes, i found the equation of the directrix. the answer in my book is $x^2+y^2-8x-8y-2xy=0$

also, the distance between the origin and (1,1) is $\sqrt{2}$

Here is a graph to help you see that it was slanted.

**c_323_h** · April 17th 2006, 07:50 AM

Originally Posted by ThePerfectHacker

I used the formula given a non-vertical line,
$Ax+By+C=0$ and a point not on the line, $(x_0,y_0)$ then the distance is,
$\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

is there a specific name for this formula.
Thanks for the graph

**TD!** · April 17th 2006, 07:54 AM

It's just a general formula for the distance, you have an equivalent one for the distance of a point to a plane, namely the analogous:

$<br /> \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}<br />$

You can derive these formulas, if you wish. I doubt they have a 'special name'.

**ThePerfectHacker** · April 17th 2006, 07:55 AM

Originally Posted by c_323_h

Is there a specific name for this equation?
Thanks for the graph.

I never heard of this one, I have seen it used a number of times.

**c_323_h** · April 18th 2006, 04:15 AM

Originally Posted by ThePerfectHacker

I used the formula given a non-vertical line,
$Ax+By+C=0$ and a point not on the line, $(x_0,y_0)$ then the distance is,
$\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

Hope this helps.

ok, so i am trying to find the distance from the focus to what? i need some clarification on this

**ThePerfectHacker** · April 18th 2006, 04:43 AM

Originally Posted by c_323_h

ok, so i am trying to find the distance from the focus to what? i need some clarification on this

THe definition of parabola is equidistant from a line (directrix) and a point (focus).
Let $(x,y)$ be a point on a parabola.
You found directrix to be $y=-x-2$
And focus is at $(1,1)$ .

Thus the distance to the focus is:
$\sqrt{(x-1)^2+(y-1)^2$
this is simply the distance between two points.

For the line you have the distance as:
$<br /> \frac{|x+y+2|}{\sqrt{2}}<br />$
Because in standard form, the equation of the directrix is $x+y+2=0$
Also remember when you find distance from a point to line you need it to be perpendicular because when we find distance from a point to a curve we keep it mimizied, and the perpendicular line to the directrix is the distance because it takes the shortest path. Thus, this is reason why we use this formula.

Now you set them equal to each other.

**c_323_h** · April 18th 2006, 06:26 AM

ok, i understand now, i was a little confused because i've never seen this before: $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

**ThePerfectHacker** · April 18th 2006, 09:05 AM

Originally Posted by c_323_h

ok, i understand now, i was a little confused because i've never seen this before: $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

The important thing is that you understand it, do you?

**c_323_h** · April 18th 2006, 09:14 AM

well, i know that $ax+by+c=0$ is the standard equation of a line and $\sqrt{a^2+b^2}$ is the distance formula (i think), but i don't know why you divide the two. maybe you can enlighten me, or even be my mentor! i googled it and try to find out how it's derived but no luck on understanding it.

i understand why you set the two equations together, because the distance from the focus to any point is equal to the distance to the directrix. i ended up getting the right answer. is there anything in math that confuses you?

**ThePerfectHacker** · April 18th 2006, 09:34 AM

Originally Posted by c_323_h

well, i know that $ax+by+c=0$ is the standard equation of a line and $\sqrt{a^2+b^2}$ is the distance formula (i think), but i don't know why you divide the two. maybe you can enlighten me, or even be my mentor! i googled it and try to find out how it's derived but no luck on understanding it.

i understand why you set the two equations together, because the distance from the focus to any point is equal to the distance to the directrix. i ended up getting the right answer. is there anything in math that confuses you?

As of know I can think of three ways to derive this formula:
1)Basic analytic geometry and facts (the long hard way).
2)The Cheap Vector method (much easier and faster)
3)The derivative (another easier method).

Which one do you prefer?
The first one is on level of understanding. The second one you need to know some properties of vectors. The third you need an understanding in calculus.
---------------
The basic idea is when we have a line,
$Ax+By+C=0$ with $A,B\not =0$ .
And a point not on a line (at point $(x_0,y_0)$ ).
Then to find the distance between the point and a line you need to draw a perpendicular line [which is possible by the parallel posutlate

- whatever igonore that] because distance needs to be minimized and that is the perpendicular line.

**c_323_h** · April 18th 2006, 09:52 AM

i'll be able to understand 1 and 2. does any part of math confuse you?

**ThePerfectHacker** · April 18th 2006, 02:54 PM

Originally Posted by c_323_h

i'll be able to understand 1 and 2. does any part of math confuse you?

Later on I will develope the formula. Cannot do this now takes to long.

If you are asking me whether I can answer any math question the answer is of course not. I do not know the whole thing. The last person to accomplish such a feat was Gauss back in the 18-19th century.

Thread: Tough Parabola

LinkBack

Thread Tools

Display

Tough Parabola

Similar Math Help Forum Discussions

here's a tough one for you

I have a tough d.e.

tough tough gauss law

Tough one

[SOLVED] tough one

Search Tags