I have an assignment for class that I just cannot figure out right now.

Find the Cartesian equation for the set of all points whose distance from the origin is twice their distance from the line y + x = 1.

What kind of conic section is this, and where are its foci?

That's the question. I realize that it's a hyperbola, and I know that you have to tilt the hyperbola I just don't know or understand how.

There is nothing in my math book as to how to do it, and searching the internet hasn't been much help either.

Could someone please help me?

## Hyperbola problem help.

**Moderators:** gmalivuk, Moderators General, Prelates

### Re: Hyperbola problem help.

Let L be the line x + y = 1.

Let P = (x,y) be a general point whose distance from the origin is twice its distance from L.

The tricky part would probably be coming up with a non-messy expression for the distance from P to L.

Let (a,b) be the point on L closest to P.

POSSIBLY HELPFUL OBSERVATION: Since the vector from (a,b) to P is perpendicular to L, that vector must have the form (k,k).

This gives (a,b) + (k,k) = (x,y), i.e. x = a + k and y = b + k. So x + y = a + b + 2k = 1 + 2k.

So k = (x + y - 1)/2.

I haven't finished working it out, but I hope this is less messy than other possible approaches.

Let P = (x,y) be a general point whose distance from the origin is twice its distance from L.

The tricky part would probably be coming up with a non-messy expression for the distance from P to L.

Let (a,b) be the point on L closest to P.

POSSIBLY HELPFUL OBSERVATION: Since the vector from (a,b) to P is perpendicular to L, that vector must have the form (k,k).

This gives (a,b) + (k,k) = (x,y), i.e. x = a + k and y = b + k. So x + y = a + b + 2k = 1 + 2k.

So k = (x + y - 1)/2.

I haven't finished working it out, but I hope this is less messy than other possible approaches.

### Re: Hyperbola problem help.

With skullturf’s observation, you’re almost done. Just plug in the distance from the origin of the point (x, y) to your required equality.

wee free kings

### Re: Hyperbola problem help.

I'm still somewhat confused though.

I don't understand how to find the equation for the hyperbola or how to rotate the axes so that the hyperbola is at an angle.

I understand somewhat what skullturf is trying to say but it is still blurry to me.

If anyone could explain this for me in a very simple manner that would be extremely appreciated.

I don't understand how to find the equation for the hyperbola or how to rotate the axes so that the hyperbola is at an angle.

I understand somewhat what skullturf is trying to say but it is still blurry to me.

If anyone could explain this for me in a very simple manner that would be extremely appreciated.

### Re: Hyperbola problem help.

Do you know an algebraic expression for the distance between two points, and/or for the length of a vector?

### Re: Hyperbola problem help.

Yes I know that the distance between any two points would be square root((x-a)^2+(y-b)^2).

### Re: Hyperbola problem help.

chewey wrote:Yes I know that the distance between any two points would be square root((x-a)^2+(y-b)^2).

And similarly, the length of any vector (u,v) is sqrt(u^2 + v^2).

### Re: Hyperbola problem help.

To get the equation you are looking for, take an arbitrary point (x,y) and then translate the phrase "distance from the origin is twice their distance from the line" into symbols.

You should get something that can be put into the form [imath]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/imath], which means that it is a conic section. Which type of conic section depends on the value of [imath]B^2-4AC[/imath]. (Does this sound familiar? Not every class teaches this.) As you say, this one turns out to be a hyperbola. Some sort of change of variable will transform the equation into the standard form [imath]\frac{x^2}{G^2}-\frac{y^2}{H^2}=\pm 1[/imath]

You should get something that can be put into the form [imath]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/imath], which means that it is a conic section. Which type of conic section depends on the value of [imath]B^2-4AC[/imath]. (Does this sound familiar? Not every class teaches this.) As you say, this one turns out to be a hyperbola. Some sort of change of variable will transform the equation into the standard form [imath]\frac{x^2}{G^2}-\frac{y^2}{H^2}=\pm 1[/imath]

### Re: Hyperbola problem help.

Once you have the equation in x and y, if you want to change coordinates to put it in standard form a good place to start would be identifying the asymptotes.

wee free kings

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