Conic Rotation - identity and write an equation of the translated or rotated graph in general form.

Question

Determine the graph identity y^2 + 8x= 0 for ϕ=π/6 and write an equation of the translated or rotated graph in general form.

Answer 1

Hi Elaine W.,

Actually a modestly challenging problem, thank you!

Suggest you first graph this function, in the non-rotated state. It's a parabola, isn't it, opening towards the negative x axis direction? (if you don't recognize, just interchange the variables: y-> x, -x->y and you will see that. OK, so how do you rotate such a beast?

A. You rotate by a linear geometric-trig based method. Draw a right triangle from the origin, longer leg on x-axis, incorporating a base angle of pi/6 (i.e., 30 degrees), so that's a 30-60-90 triangle. The hypotenuse "direction" of that triangle, which takes the functional place of the axis for the variable "x" in the rotated parabola equation, includes scaled contributions from the previous x and y sides, doesn't it; you can write that as a function of the former x and y coordinates (Pythagorean theorem, and mind the scaling factor in the original equation!)

Similarly, consider a 30-60-90 triangle with a longer leg along the former "y" axis, and opening 30 degrees positive: thus the short leg represents the former "-x" direction.The hypotenuse (sitting at an angle of 2*pi/3 from the former x axis), is the new "y" coordinate for your transformed equation, and includes scaled contributions of the previous x and y, according to the Pythagorean theorem.

So now you have expressions for the "new" x and "new" y in a parabola equation; you substitute in those expressions in place of the terms of the old x and y coordinates, and crank!

You should expect that you will have cross-terms whenever you rotate a conic section in this way: not just y^2 and x, but also xy and maybe other beasts. Unfortunately, not every equation that has x, y, xy, x^2, and y^2 terms in it is a rotated parabola (i.e. it doesn't work the other way!).

That's in Cartesian coordinates, anyway. I would imagine that you could transform to polar coordinates initially, and then just tweak the .phi. value. But, that's up to you. Polar would be more elegant, don't you think?

Oh and also, you may need to give some thought to the 2nd "limb" of the parabola; it may have slightly different form (particularly for signs of terms) or domain restrictions from the first limb solutions you found initially.

-- Cheers, -- Mr. d.

-- By the way, in this case the and y interchange had fairly simple calculations; in general, it's just a trigonometric derived form for the new variables and the rotation angle (sines and cosines, oh my!).