Question

Which one of the following equations represents the 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

The correct option is B $x^2-2=2\cos t;y=4{\cos }^2\frac{t}{2}$

Option A,

$x=3\cos t,y=4\sin t$

$\Rightarrow y^2=16{\sin }^{2}t$
$=16(1-{\cos }^{2}t)=16(1-\frac{x^2}{9})$
$\Rightarrow \frac{x^2}{9}+\frac{y^2}{16}=1$ which is equation of ellipse.

Option B,

$x^2-2=2\cos t,y=4{\cos }^2\frac{t}{2}$
$\Rightarrow x^2=2+2\cos t$
$=2(1+\cos t)=4{\cos }^2\frac{t}{2}=y$
$\Rightarrow x^2=y$ which is the equation of a parabola.

Option C,

$\sqrt{x}=\tan t,\sqrt{y}=\sec t$
$\Rightarrow y={\sec }^{2}t=1+{\tan }^{2}t=1+x$
$\Rightarrow y-x=1$, which is a straight line.

Option D,

$x=\sqrt{1-\sin t},y=\sin \frac{t}{2}+\cos \frac{t}{2}$
$\Rightarrow x^2=1-\sin t$

$\Rightarrow y^2={\sin }^2\frac{t}{2}+{\cos }^2\frac{t}{2}+2\sin \frac{t}{2}\cos \frac{t}{2}=1+\sin t$

$\Rightarrow x^2+y^2=2$, which is a circle.