"iso" means "same" so "isotherms" are lines of constant temperature.

To take your second question first, it should be obvious that xy= constant is a hyperbola: it has the x and y axes as asymptotes.

If you learned only that a hyperbola must be of the form then you should also have learned that this formula is valid only for hyperbolas with foci on the x-axis: "standard position". Quadratic formulas with "xy" terms are conic sections with axes rotated away from the x and y axes. In this case, if we take new axes at 45 degree angles to the x and y axes, that is, take x'= x- y and y'= x+ y, so that, adding the two equations, 2x= x'+ y' so x= (1/2)x'+ (1/2)y', and, subtracting, 2y= y'- x' so y= (1/2)y'- (1/2)y', then xy= (1/4)(x'+ y')(x'- y') we get or which is clearly a hyperbola.

For your first problem, again T is a constant so you have . Multiply both sides by and divide by T: or . Complete the square in the y terms: divided by 2 is and the square is [math\frac{1}{4T^2}[/tex]:

so .

That is a circle with center at and radius so it is tangent to the x-axis at (0,0).

is clearly a hyperbola. Complete the square in y again: so adding 16 to both sides give .