MCQ
Which one of the following equations represented parametrically, represents equation to a parabolic profile ?
  • A
    $x = 3 cos t ; y = 4 sin t$
  • $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

Correct option: B.
$x^2 - 2 = - 2 cos t ; y = 4 cos^2 $ $\frac{t}{2}$
b
option $( A ) x=3 \cos t: y=4 \sin t$

eliminating 't' we get, $\frac{x^{2}}{9}+\frac{y^{2}}{16}=1$ Clearly this is an ellipse.

option

(B) $x^{2}=-\cos t: y=4 \cos ^{2} \frac{t}{2}=2(1+\cos t)$

eliminating 't' we get, $x^{2}=-\left(\frac{y}{2}-1\right)$ Clearly this is parabola.

option (C) $\sqrt{x}=\tan t \sqrt{y}=\sec t$

eliminating 't' we get, $y-x=1$ Clearly this is a straight line

option (D) $x=\sqrt{1-\sin t}: y=\sin \frac{t}{2}+\cos \frac{t}{2}$

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

eliminating 't" we get, $x^{2}+y^{2}=2$ Clearly this is a circle.

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