Question

Which of the following represents a parabola?

• A. \(x=4\cos t,\ y=4\sin t\)
• B. \(x^2-2=-2\cos t,\ y=\cos^2\left(\frac t2\right)\)
• C. \(\sqrt{x}=\tan t,\ \sqrt{y}=\sec t\)
• D. \(x=\sqrt{1-\sin t},\ y=\sin\left(\frac t2\right)+\cos\left(\frac t2\right)\)

Hint:

To spot a parabola, eliminate the parameter \(t\) and reduce each pair to a relation between \(x\) and \(y\). The conic is a parabola only when the result has exactly one squared variable, i.e. it simplifies to the form \(y=ax^2+bx+c\) (or \(x=ay^2+by+c\)).

Answer 1

The Correct Option is B

Step 1: Check option 1.
$x=4\cos t$, $y=4\sin t$: $x^2+y^2=16$, a circle.
Step 2: Set up option 2 with the identity $\cos t=2\cos^2(t/2)-1$.
$y=\cos^2(t/2) \implies \cos t=2y-1$.
Step 3: Eliminate $t$ in option 2.
$x^2-2=-2\cos t=-2(2y-1)=-4y+2 \implies x^2=-4y+4=-4(y-1)$. This is a parabola.
Step 4: Check option 3.
$\sqrt{x}=\tan t$, $\sqrt{y}=\sec t$: $y-x=1$, a line.
Step 5: Check option 4.
$y^2=1+\sin t$, $x^2=1-\sin t$: $x^2+y^2=2$, a circle.
Step 6: Conclude.
Only option 2 represents a parabola. \[ \boxed{x^2=-4(y-1)} \]