Hi,

How do we solve systems of conic equations? For example:

$\displaystyle \frac{(x-3)^2}{16}-\frac{(y+9)^2}{64}=1$

and

$\displaystyle \frac{(x-9)^2}{9}+\frac{(y+2)^2}{49}=1$

How do we find the point(s) of intersection?

Thanks!

Printable View

- Mar 4th 2011, 01:50 PMmasougSolving Conic Equations
Hi,

How do we solve systems of conic equations? For example:

$\displaystyle \frac{(x-3)^2}{16}-\frac{(y+9)^2}{64}=1$

and

$\displaystyle \frac{(x-9)^2}{9}+\frac{(y+2)^2}{49}=1$

How do we find the point(s) of intersection?

Thanks! - Mar 4th 2011, 02:29 PMTKHunny
Not sure I've ever heard the term "Conic Equation". Anyway...

First, you should come to grips with the fact that there may be no point of intersection.

After that, a little quick homework can narrow things down quickly. The first, centered at (3,-9) and reaching 4 or 8, depending which way you go, might suggest that it does run into the second, whcih is centered at (9,-2) and reaches 3 or 7, depending on direction.

Next, that ellipse doesn't get anywhere near Quadrant II, so we can just ignore that piece of the hyperbola.

Hyperbola's asymptotes are pretty clear at

y = 2x+3 and y = 15-2x

Again, the ellipse doesn't get all that close to y = 2x+3, so I'm guessing that a new problem is in order. Find the points of intersection between the ellipse and y = 15-2x. This WILL tell you where to start looking. One should be closer than the other, it is an asymptote.

Now what? - Mar 4th 2011, 10:31 PMmasoug
- Mar 5th 2011, 12:57 AMHallsofIvy
What TKHunny is saying is that

**how**you determine a solution will depend strongly upon the specific conics. For these specific equations one obvioius method is to solve the first equation for y:

$\displaystyle (y+ 9)^2= 64(\frac{(x-3)^2}{16}- 1$ so

$\displaystyle y= -9\pm 8\sqrt{\frac{(x-3)^2}{16}- 1$

Now replace y in the second equation by that. You will eventually get a fourth degree equation.

Corrected thanks to Wilmer. - Mar 5th 2011, 03:02 AMWilmer
- Mar 5th 2011, 07:18 PMWilmer
Did you realise that those are 2 ellipses, IF you change 1st equation to "positive"?

The basic equation of an ellipse is x^2 / a^2 + y^2 / b^2 = 1 ; a > b

If you're "learning", why are you starting with a complicated case?

Did you google"ellipse equation"? Did you graph an ellipse to "see"?

Good "lessons" here:

http://www.solitaryroad.com/c428.html - Mar 5th 2011, 08:40 PMTKHunny
I sincerely hope I did not say that. One needs a sense of it. Unfortunately, for the discussion above, I seem to have graphed badly and put the ellipse in the wrong place. Such is one of the dangers of graphing. Once you SEE it, even if it is wrong, it is difficult to get out of your head.

Solving such problems is one of the joys of pre-calculus mathematics. I recommend finding ways to solve without the super-duper brute force suggested by HallsofIvy. For the record, this would also be my first attempt, generally, but this one seems like we can do something else with it. There's just too much symmetry to let it go.

I would try this, first, now that I have abandoned my dear old friend, Brute Force.

Solve the ellipse for (x-3)^2 and substitute into the hyperbola. See if you can get it. (x-3)^2 = 12x - (1/49)[3123+36y+9y^2].

This substitution turns the hyperbola into: 2352x = 19597 + 1026y + 85y^2

If you also solve the hyperbola for (y+2)^2, getting (y+2)^2 = 4x^2 - 24x - 14y -105

and make a similar substitution into the ellipse, the ellipse reduces to 126y = 2583 - 1098x + 85x^2

Now, it's almost easy. Who can't find the intersections of two parabolas? - Mar 6th 2011, 12:06 AMWilmer
To really get familiar with "intersecting points", I suggest you try 2 ellipses; as example:

x^2 / 36 + y^2 / 9 = 1 (horizontal)

x^2 / 4 + y^2 / 25 = 1 (vertical)

You should end up with 4 intersecting points:

[4/sqrt(6), 5/sqrt(3)], [-4/sqrt(6), 5/sqrt(3)], [-4/sqrt(6), -5/sqrt(3)], [4/sqrt(6), -5/sqrt(3)]

And you can actually "have fun" doing the work!! - Mar 6th 2011, 06:30 PMmasoug
Thanks for the helpful replies!

I can see how this works now! - Sep 29th 2012, 03:38 PMZeroDivisionErrorRe: Solving Conic Equations
The first equation is a hyperbola. I'm going to assume multiplying both sides by -1 is what you mean by "[changing] the 1st equation to 'positive'". Even if you multiply the equation by -1, you still won't get an ellipse, because the 1 on the right side will become negative.

- Sep 29th 2012, 03:53 PMWilmerRe: Solving Conic Equations