Question

Which one of the following equations represent parametric equation to a parabolic curve?

• A. $x=3 \cos t ; y=4 \sin t$
• B. $x^{2}-2=2 \cos t ; y=4 \cos ^{2} \frac{t}{2}$
• C. $\sqrt{x}=\tan t ; \sqrt{y}=\sec t$
• D. $x=\sqrt{1-\sin t} ; y=\sin \frac{t}{2}+\cos \frac{t}{2}$

Answer 1

Answer: (B)

$x=3 \cos t, y=4 \sin t$
Eliminating $t$, we have
$\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$
which is an ellipse. Therefore,
$x^{2}-2=2 \cos t $ and $ y=4 \cos ^{2} \frac{t}{2}$
or $y=2(1+2 \cos t)$
and $y=2\left(1+\frac{x^{2}-2}{2}\right)$
which is a parabola.
$\sqrt{x}=\tan t ; \sqrt{y}=\sec t$
Eliminating $t$, we have
$y-x=1$
which is a straight line.
$x=\sqrt{1-\sin t}$
$y=\sin \frac{t}{2}+\cos \frac{t}{2}$
Eliminating $t$, we have
$x^{2}+y^{2}=1-\sin t+1+\sin t=2$
which is a circle.