# Tough vectors question...

• Feb 7th 2011, 10:39 PM
ice_syncer
Tough vectors question...
Let r be a postition vector of a viariable point in cartesian plane OXY such that
r.(10j - 8i - r/|r| ) = 40 and p1 = max{(|r + 2i - 3j|)^2} , p2 = min{(|r + 2i - 3j|)^2}
. A tanget line is drawn to the curve y = 8/(x^2) at the point A with abcissa 2. The drawn line cuts x-axis at a point B.
Now answer the following questions based on above paragraph

Q1 p2 is equal to ?

Q2 p1+p2 = ?

Q3 vector AB . vector OB = ?

please explain exactly what is vector r ??
• Feb 8th 2011, 01:01 AM
Ackbeet
Let's say that $\displaystyle \mathbf{r}=\langle r_{1},r_{2}\rangle.$ If you plug this into the condition

$\displaystyle \mathbf{r}\cdot\left(\langle -8,10\rangle-\dfrac{\mathbf{r}}{\|\mathbf{r}\|}\right)=40,$

you'll end up with the equation for an ellipse in the coordinates $\displaystyle \langle r_{1},r_{2}\rangle.$

You can verify that the equation for the ellipse is the following:

$\displaystyle 63r_{1}^{2}+99r_{2}^{2}-160r_{1}r_{2}+640r_{1}-800r_{2}+1600=0.$

So the answer to your question of "what is $\displaystyle \mathbf{r}?$" is this: $\displaystyle \mathbf{r}$ is the coordinate for a point on the ellipse described by the above equation.

Does that help?
• Feb 8th 2011, 09:49 AM
ice_syncer
how do you know that it is an ellipse ?
• Feb 8th 2011, 09:50 AM
ice_syncer
I mean... how did you come to know that it is an ellipse at first sight..
• Feb 8th 2011, 09:55 AM
Ackbeet
Quote:

Originally Posted by ice_syncer
I mean... how did you come to know that it is an ellipse at first sight..

I didn't. I recognized it as an ellipse after I worked out the equation in post # 2.
• Feb 8th 2011, 10:04 AM
ice_syncer
Ah ok.. Now I get it... so to find p2 or p1 we jsut substitute r as <x,y> ... dot product it with itself, then d/dx the expression and equate d/dx = 0, find the maxima or minima depending on p1 or p2's demand .... the n find y from the ellipse equation... nice.. if you have an easier way please suggest. Thanks for making me understand what vector r is.
• Feb 8th 2011, 10:30 AM
Ackbeet
I would highly recommend drawing a picture. Drawing a picture enabled me to conclude that, after all, it's not an ellipse. If you look here, you will see that you need the discriminant to be negative, which is not true in this case. The graph looks something like an hyperbola, but I don't think it's an hyperbola. The graph to which I've linked is centered about the point (-2,3), which figures largely in your expressions for p1 and p2. I do think that p2 is well-defined. The point (-2,3) lies just above the upper branch of the relation in question, and finding the minimum distance from that point to the graph should be doable, at least theoretically. However, considering that the graph appears to go on to infinity in multiple directions, I don't think your p1 is well-defined.