Let a be a fixed positive real number with a≠1 . Show that…

Let $a$ be a fixed positive real number with $a\neq1$. Show that as $t$ ranges over $[0,2\pi)$, the curve $\ z=ae^{it}+\dfrac1{ae^{it}}\ $ traces an ellipse.

Step 1 · Simplify $\dfrac1{ae^{it}}$

$$ \frac1{ae^{it}}=\frac1a\,e^{-it} $$

so:

$$ z=ae^{it}+\frac1a e^{-it} $$

Step 2 · Expand via Euler's formula

$$ z=a(\cos t+i\sin t)+\frac1a(\cos t-i\sin t)=\left(a+\frac1a\right)\cos t+i\left(a-\frac1a\right)\sin t $$

Step 3 · Identify real and imaginary parts

Writing $z=x+iy$:

$$ x=\left(a+\frac1a\right)\cos t, \qquad y=\left(a-\frac1a\right)\sin t $$

Step 4 · Eliminate the parameter $t$

Since $a\neq1$ (and $a>0$), both $a+\frac1a\neq0$ and $a-\frac1a\neq0$, so we may divide:

$$ \frac{x}{a+\frac1a}=\cos t, \qquad \frac{y}{a-\frac1a}=\sin t $$

Squaring and adding, using $\cos^2t+\sin^2t=1$:

$$ \boxed{\ \frac{x^2}{\left(a+\frac1a\right)^2}+\frac{y^2}{\left(a-\frac1a\right)^2}=1\ } $$

Step 5 · Identify the ellipse

This is the standard equation of an ellipse centered at the origin, with semi-axes $A=a+\dfrac1a$ (along the real axis) and $B=\left|a-\dfrac1a\right|$ (along the imaginary axis). Since $a+\frac1a>\left|a-\frac1a\right|$ for all $a>0,\ a\neq1$ (as $A^2-B^2=4>0$), the major axis always lies along the real axis, and the foci are located at $z=\pm2$ (since $A^2-B^2=\left(a+\frac1a\right)^2-\left(a-\frac1a\right)^2=4$, giving focal distance $c=2$, independent of $a$).

Verification

For $a=2$: $x=2.5\cos t$, $y=1.5\sin t$, and indeed $\frac{x^2}{2.5^2}+\frac{y^2}{1.5^2}=\cos^2t+\sin^2t=1$ for every $t$. The relation $A^2-B^2=2.5^2-1.5^2=6.25-2.25=4$ also checks out.