Question

What is the rectangular form of the curve described by the parametric equations: `x = acostheta+bsintheta` and `y = bcostheta+asintheta` ?

Answer 1

Given
`x=acostheta + bsintheta.......[1]`

`y=bcostheta + asintheta........[2]`

From {1] we get

`costheta+b/asin theta=x/a....[5]`

From [2] we get

`costheta+a/bsin theta=y/b...[6]`

subtracting [6] from [5]

`sintheta(b/a-a/b)=(x/a-y/b)`

`sintheta=(bx-ay)/(b^2-a^2)...[7]`

similarly from [1]

`a/bcostheta+sintheta=x/b....[8]`

and from [2]

`b/acostheta+sintheta=y/a....[9]`

subtracting [9] from [8] we get

`costheta(a/b-b/a)=x/b-y/a`

`costheta=(ax-by)/(a^2-b^2)....[10]`

From [7] and [10] we get

`((bx-ay)/(b^2-a^2))^2+((ax-by)/(a^2-b^2))^2=sin^2theta+cos^2theta=1`

`=>b^2x^2+a^2y^2+a^2x^2+b^2y^2-4abxy=(a^2-b^2)^2`

`=>(x^2+y^2)(a^2+b^2)-4abxy=(a^2-b^2)^2`

Answer 2

`:. {(x+y)(a-b)}^2+{(x-y)(a+b)}^2=2(a^2-b^2)^2.`

Explanation:

We rewrite the given eqns. as,

`acostheta+bsintheta=x, and, bcostheta+asintheta=y.`

Solving these for `costheta, and, sintheta,` we get,

`2costheta=(x+y)/(a+b)+(x-y)/(a-b),`

and,

`2sintheta=(x+y)/(a+b)-(x-y)/(a-b).`

`:. (2costheta)^2+(2sintheta)^2=2{((x+y)/(a+b))^2+((x-y)/(a-b))^2}.`

`:. {(x+y)(a-b)}^2+{(x-y)(a+b)}^2=2(a^2-b^2)^2.`

Answer 3

`(a^2+b^2)x^2+(a^2+b^2)y^2-4abxy-(a^2-b^2)^2 = 0`

Explanation:

Yet another way...

Given:

`x = acostheta+bsintheta`

`y = bcostheta+asintheta`

Then:

`xy = (acostheta+bsintheta)(bcostheta+asintheta)`

`color(white)(xy) = ab(cos^2theta+sin^2theta)+(a^2+b^2)costhetasintheta`

`color(white)(xy) = ab+(a^2+b^2)costhetasintheta`

So:

`costhetasintheta = (xy-ab)/(a^2+b^2)`

Also we find:

`x^2+y^2 = (acostheta+bsintheta)^2+(bcostheta+asintheta)^2`

`color(white)(x^2+y^2) = (a^2+b^2)(cos^2theta+sin^2theta)+4abcosthetasintheta`

`color(white)(x^2+y^2) = (a^2+b^2)+4abcosthetasintheta`

So:

`costhetasintheta = (x^2+y^2-(a^2+b^2))/(4ab)`

Putting these together, we find:

`(xy-ab)/(a^2+b^2) = (x^2+y^2-(a^2+b^2))/(4ab)`

Cross multiplying, this becomes:

`4ab(xy-ab) = (a^2+b^2)(x^2+y^2-(a^2+b^2))`

Hence:

`0 = (a^2+b^2)x^2+(a^2+b^2)y^2-4abxy-((a^2+b^2)^2-4a^2b^2)`

`color(white)(0) = (a^2+b^2)x^2+(a^2+b^2)y^2-4abxy-((a^4+2a^2b^2+b^4)-4a^2b^2)`

`color(white)(0) = (a^2+b^2)x^2+(a^2+b^2)y^2-4abxy-(a^4-2a^2b^2+b^4)`

`color(white)(0) = (a^2+b^2)x^2+(a^2+b^2)y^2-4abxy-(a^2-b^2)^2`

Notes

We can perform some simple checks on the answer by trying various combinations of `a` and `b`

For example, with `a=b` the original equations give us:

`x = acos theta + asin theta = y`

and the derived equation gives us:

`0 = (a^2+a^2)x^2+(a^2+a^2)y^2-4a^2xy-(a^2-a^2)^2`

`color(white)(0) = 2a^2(x^2+y^2-2xy)`

`color(white)(0) = 2a^2(x-y)^2`

hence `x = y`

For another example, with `b=0` the original equations give us:

`x = acos theta`

`y = asin theta`

which is the parametric form of a circle with radius `a` centred on the origin.

The derived equation gives us:

`0 = a^2x^2+a^2y^2-(a^2)^2 = a^2(x^2+y^2-a^2)`

That is:

`x^2+y^2 = a^2`

which is the standard form of the equation of a circle with radius `a` centred on the origin.