Describe all points (x,y) satisfying the equation

[AKJ1]

Homework Statement

Problem that was gone over in class, but I missed part of the lecture

Describe all points (x,y) satisfying | p - s | + | p - t | = Q

where,
Vector p = <x,y>
Vector s = <a,b>
Vector t = <c,d>

a,b,c,d and Q are constants and Q is greater than | s - t |

The Attempt at a Solution

I think the graph is a cylinder, but I am having a lot of trouble trying to verify or describe that. I've tried writing in the components and brute forcing it, but it ends up quite hairy. I also have tried squaring both sides and using the fact that if u and v are vectors then

| u + v | ^2 = |u|^2 + |v|^2

But I haven't made much progress there either. Is there some identity or simplification I am not seeing?

 

[blue_leaf77]

Have you learned conic sections? If you have, I suggest that you review again your notes especially about how each type of curve resulting from various conic sections is geometrically defined. If you have not, then you might have to go through lines of tedious algebra, although not that lengthy.

 

[AKJ1]

Have you learned conic sections? If you have, I suggest that you review again your notes especially about how each type of curve resulting from various conic sections is geometrically defined. If you have not, then you might have to go through lines of tedious algebra, although not that lengthy.

http://www.stewartcalculus.com/data/ESSENTIAL CALCULUS Early Transcendentals/upfiles/ess-reviewofconics.pdf

Thats about the extent of my knowledge of conics. Which is probably sufficient, but I am having a hard time making the connection.

sqrt( (x-a)^2 + (y-b)^2 ) = Q - sqrt( (x-c)^2 + (y-d)^2 )

square both sides

x^2 - 2xa + a^2 + y^2 -2yb + b^2 = Q^2 -2Q*sqrt( (x-c)^2 + (y-d)^2 ) + x^2 -2xc + c^2 + y^2 - 2yd + d^2

collect some terms and simplify

2x(c-a) + 2y(d-b) + a^2 + b^2 - c^2 - d^2 - Q^2 = -2Q*sqrt( (x-c)^2 + (y-d)^2 )

About here is where I get stuck. I know I was given information about Q, but I am not quite sure how to take advantage of that just yet.

 

[blue_leaf77]

http://www.stewartcalculus.com/data/ESSENTIAL CALCULUS Early Transcendentals/upfiles/ess-reviewofconics.pdf

Thats about the extent of my knowledge of conics. Which is probably sufficient, but I am having a hard time making the connection.

sqrt( (x-a)^2 + (y-b)^2 ) = k - sqrt( (x-c)^2 + (y-d)^2 )

square both sides

x^2 - 2xa + a^2 + y^2 -2yb + b^2 = k^2 -2k*sqrt( (x-c)^2 + (y-d)^2 ) + x^2 -2xc + c^2 + y^2 - 2yd + d^2

collect some terms and simplify

2x(c-a) + 2y(d-b) + a^2 + b^2 - c^2 - d^2 - k^2 = -2k*sqrt( (x-c)^2 + (y-d)^2 )

About here is where I get stuck. I know I was given information about K, but I am not quite sure how to take advantage of that just yet.

Take one more square. Obviously, it's going to be a tedious work though, may be there are other members who can propose quicker method. But if I were you, I will try coordinate transformation first. For example bring the origin to one of the fixed point, for example the vector $\mathbf{t}$, and then rotate the translated coordinate such that the vector $\mathbf{s}$ lies in the new x axis. In short, after applying this transformation you get
$$
\begin{aligned}
\mathbf{t} &\rightarrow 0 \\
\mathbf{s} &\rightarrow \gamma\hat{\mathbf{x}}'
\end{aligned}
$$.
where $(x',y')$ are the new coordinate system and $\gamma = |\mathbf{s}-\mathbf{t}|$.

 

[AKJ1]

Take one more square. Obviously, it's going to be a tedious work though, may be there are other members who can propose quicker method. But if I were you, I will try coordinate transformation first. For example bring the origin to one of the fixed point, for example the vector $\mathbf{t}$, and then rotate the translated coordinate such that the vector $\mathbf{s}$ lies in the new x axis. In short, after applying this transformation you get
$$
\begin{aligned}
\mathbf{t} &\rightarrow 0 \\
\mathbf{s} &\rightarrow \gamma\hat{\mathbf{x}}'
\end{aligned}
$$.
where $(x',y')$ are the new coordinate system and $\gamma = |\mathbf{s}-\mathbf{t}|$.

Will give it a go, thanks! :)

 

[Mark44]

Homework Statement

Problem that was gone over in class, but I missed part of the lecture

Describe all points (x,y) satisfying | p - s | + | p - t | = Q

Note that the above is an equation, not an expression, as implied in your thread title ("Describe all points (x,y) satisfying the expression"). If a point (x, y) satisfies an equation, the equation becomes a true statement for those values of x and y. You can't say the same thing about an expression.

I'm nitpicking here, but it is important to understand the difference between an expression and an equation.

where,
Vector p = <x,y>
Vector s = <a,b>
Vector t = <c,d>

a,b,c,d and Q are constants and Q is greater than | s - t |

The Attempt at a Solution

I think the graph is a cylinder, but I am having a lot of trouble trying to verify or describe that. I've tried writing in the components and brute forcing it, but it ends up quite hairy. I also have tried squaring both sides and using the fact that if u and v are vectors then

| u + v | ^2 = |u|^2 + |v|^2

This isn't true in general. However, it is true that $|u + v|^2 \le |u|^2 + |v|^2$.

But I haven't made much progress there either. Is there some identity or simplification I am not seeing?

 

[Ray Vickson]

http://www.stewartcalculus.com/data/ESSENTIAL CALCULUS Early Transcendentals/upfiles/ess-reviewofconics.pdf

Thats about the extent of my knowledge of conics. Which is probably sufficient, but I am having a hard time making the connection.

sqrt( (x-a)^2 + (y-b)^2 ) = Q - sqrt( (x-c)^2 + (y-d)^2 )

square both sides

x^2 - 2xa + a^2 + y^2 -2yb + b^2 = Q^2 -2Q*sqrt( (x-c)^2 + (y-d)^2 ) + x^2 -2xc + c^2 + y^2 - 2yd + d^2

collect some terms and simplify

2x(c-a) + 2y(d-b) + a^2 + b^2 - c^2 - d^2 - Q^2 = -2Q*sqrt( (x-c)^2 + (y-d)^2 )

About here is where I get stuck. I know I was given information about Q, but I am not quite sure how to take advantage of that just yet.

You must have missed the part in your link that told you exactly what is the solution to your problem.

 

[WWGD]

Take one more square..

Maybe s/he can't spare a square ;).

 

[WWGD]

Will give it a go, thanks! :)

I think what Blue Leaf was trying to get at is that the set of points you describe is the one whose sum of distances to two points is fixed. This is a standard, well-known space.

 

[blue_leaf77]

I think what Blue Leaf was trying to get at is that the set of points you describe is the one whose sum of distances to two points is fixed. This is a standard, well-known space.

Yep exactly, that's why I brought up straight away the subject of conic section to narrow down the possibility to a few curves. But we have no idea though, what the problem exactly ask, whether it can be answered by explanation by words or must be provided with mathematical steps.