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).

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- Apr 16th 2006, 05:26 PMc_323_hTough Parabola
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). - Apr 16th 2006, 06:35 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

Since the vertex is at $\displaystyle (0,0)$ and focus $\displaystyle (1,1)$ then, drawing a line between them and extending it in the opposite direction the same distance you end up at $\displaystyle (-1,-1)$. That means the directrix passes through this point. Furthermore, it is perpendicular to the line you drawn. Thus, its slope is $\displaystyle -1$ using slope and intercept-formula we have,

$\displaystyle y+1=-1(x+1)$

Thus,

$\displaystyle y=-x-2$ is equation of directrix.

Next, let $\displaystyle (x,y)$ 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,

$\displaystyle \sqrt{(x-1)^2+(y-1)^2}=\frac{|x+y+2|}{\sqrt{2}}$

Do not worry the abosolute value disappers when you square both sides.

-------

I used the formula given a non-vertical line,

$\displaystyle Ax+By+C=0$ and a point not on the line, $\displaystyle (x_0,y_0)$ then the distance is,

$\displaystyle \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

Hope this helps. - Apr 16th 2006, 06:55 PMc_323_h
yes, i found the equation of the directrix. the answer in my book is $\displaystyle x^2+y^2-8x-8y-2xy=0$

also, the distance between the origin and (1,1) is $\displaystyle \sqrt{2}$ - Apr 17th 2006, 07:36 AMThePerfectHackerQuote:

Originally Posted by**c_323_h**

- Apr 17th 2006, 07:50 AMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

Thanks for the graph - Apr 17th 2006, 07:54 AMTD!
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:

$\displaystyle

\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}

$

You can derive these formulas, if you wish. I doubt they have a 'special name'. - Apr 17th 2006, 07:55 AMThePerfectHackerQuote:

Originally Posted by**c_323_h**

- Apr 18th 2006, 04:15 AMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

- Apr 18th 2006, 04:43 AMThePerfectHackerQuote:

Originally Posted by**c_323_h**

Let $\displaystyle (x,y)$ be a point on a parabola.

You found directrix to be $\displaystyle y=-x-2$

And focus is at $\displaystyle (1,1)$.

Thus the distance to the focus is:

$\displaystyle \sqrt{(x-1)^2+(y-1)^2$

this is simply the distance between two points.

For the line you have the distance as:

$\displaystyle

\frac{|x+y+2|}{\sqrt{2}}

$

Because in standard form, the equation of the directrix is $\displaystyle x+y+2=0$

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. - Apr 18th 2006, 06:26 AMc_323_h
ok, i understand now, i was a little confused because i've never seen this before: $\displaystyle \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$

- Apr 18th 2006, 09:05 AMThePerfectHackerQuote:

Originally Posted by**c_323_h**

- Apr 18th 2006, 09:14 AMc_323_h
well, i know that $\displaystyle ax+by+c=0$ is the standard equation of a line and $\displaystyle \sqrt{a^2+b^2}$ 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? - Apr 18th 2006, 09:34 AMThePerfectHackerQuote:

Originally Posted by**c_323_h**

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,

$\displaystyle Ax+By+C=0$ with $\displaystyle A,B\not =0$.

And a point not on a line (at point $\displaystyle (x_0,y_0)$).

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 :D - whatever igonore that] because distance needs to be minimized and that is the perpendicular line. - Apr 18th 2006, 09:52 AMc_323_h
i'll be able to understand 1 and 2. does any part of math confuse you?

- Apr 18th 2006, 02:54 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

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.