maths question

Martyn_Thomas

Default maths question
some1 help me plz im stuck

sketch the relation:

TX123

u get an equation for the circle. Multiply it all out to get

x^2 - 4y2 -10x-8y+65=0

the circle centre is (5,4)

Michael Collins

It isn't actually the equation for a circle as it hasn't got equal coefficients infront of the x & y variables. Normally this would yeild an ellipse, but because the constant on the RHS is negative & the coefficient of y is too, it won't even be that!

At a guess it'll probably be two curves one "u" shaped in the top half of the plane, and one "n" shaped in the bottom.

It looks as if the OP has been banned so there is no point in asking where this unusual function came from is there!?

TX123

a circle does not need equal coefficiants on both the x and y

MathsManiac

No, but it needs equal coefficients in front of x^2 and y^2, which I suspect is what MC meant.

What you've got is a hyperbola. Attached is its plot.

Martyns_Curve.gif

It's a conic section, like the circle, ellipse and parabola. If you want to see one in real life, it's the shape of the shadow cast on the wall by a lamp with a cylindirical lampshade.

TX123

ah i see. silly me. good graph. what software did u use?

Michael Collins

MathsManiac wrote:

No, but it needs equal coefficients in front of x^2 and y^2, which I suspect is what MC meant.

Yep, that's what I meant.

MathsManiac

TX123 wrote:

ah i see. silly me. good graph. what software did u use?

Maple

MathsManiac

It occurs to me that giving a computer generated plot is not especially helpful to someone trying to figure out how to sketch such a curve. So here's some extra help for the silenced Mr T:

First of all, one should be able to spot that this is a translated version of the curve XX/36 - YY/9 = -1, (under the translation (0,0) to/from (5,1) ).

You then need to recognise that this has the structure of a conic centred at the origin and oriented to the axes: multiple of x^2 and multiple of y^2 added to give a constant. It's more usual for the constant to be treated as positive, so change signs: -XX/36 + YY/9 = 1.

If the coefficiants of X^2 and Y^2 are equal, it's a circle.
If they're not equal, then: it's an ellipse if they have the same sign, and a hyperbola if they have opposite sign. [If it had been an ellipse, you could sketch it by locating the top, bottom, left- and right-most points.]

For a hyperbola, if the minus is in front of the X^2, then the branches are top and bottom; if it's in front of the Y^2, they're left and right.

The asymptotes of the hyperbola -XX/aa + YY/bb = 1 are X/a + Y/b = 0 and X/a - Y/b = 0.

So the asymptotes of -XX/36 + YY/9 = 1 are -X/6 + Y/3 = 0 and -X/6 - Y/3 = 0, which can be written as -X+2Y=0 and -X-2Y=0.

This is enough to sketch the version of the curve centred at the origin, especially if you also note that -XX/36 + YY/9 = 1 passes through (0,3) and (0,-3). (These are the closest points to the origin.)

Knowing all this, you can translate the whole lot back from centre (0,0) to centre (5,1):
- It's a hyperbola with branches top and bottom;
- Its centre (of symmetry) is (5,1)
- Its asymptotes are x+2y=7 and x-2y=3
- The lowest point of the top branch is (5,4)
- The highest point of the bottom branch is (5,-1).

Hence, you get the sketch I posted earlier.