Which one of the following equations represents parametrically, parabolic profile ?

Options

A. x = 3 cost , y = 4 sint
B. `x^(2)-2=-cost,y=4cos^(2)(t/2)`
C. `sqrtx=tant,sqrty=sect`
D. `x=sqrt(1-sint),y=sin(t/2)+cos(t/2)`

Answer

To determine which of the given equations represents a parabolic profile, we will analyze each option step by step.

### Step 1: Analyze Option A
**Given:**
- \( x = 3 \cos t \)
- \( y = 4 \sin t \)

**Solution:**
1. From \( x = 3 \cos t \), we can express \( \cos t \) as:
\[ \cos t = \frac{x}{3} \]
2. From \( y = 4 \sin t \), we express \( \sin t \) as:
\[ \sin t = \frac{y}{4} \]
3. Now, squaring both equations:
\[ \cos^2 t = \left(\frac{x}{3}\right)^2 = \frac{x^2}{9} \]
\[ \sin^2 t = \left(\frac{y}{4}\right)^2 = \frac{y^2}{16} \]
4. Using the identity \( \cos^2 t + \sin^2 t = 1 \):
\[ \frac{x^2}{9} + \frac{y^2}{16} = 1 \]
5. This is the equation of an ellipse, not a parabola.

**Conclusion for Option A:** Not a parabola.

### Step 2: Analyze Option B
**Given:**
- \( x^2 - 2 = -\cos t \)
- \( y = 4 \cos^2 \frac{t}{2} \)

**Solution:**
1. Rearranging the first equation:
\[ \cos t = - (x^2 - 2) = 2 - x^2 \]
2. Using the double angle formula \( \cos t = 2 \cos^2 \frac{t}{2} - 1 \):
\[ 2 \cos^2 \frac{t}{2} - 1 = 2 - x^2 \]
\[ 2 \cos^2 \frac{t}{2} = 3 - x^2 \]
\[ \cos^2 \frac{t}{2} = \frac{3 - x^2}{2} \]
3. Now substituting into \( y \):
\[ y = 4 \left(\frac{3 - x^2}{2}\right) = 6 - 2x^2 \]
4. Rearranging gives:
\[ 2x^2 + y = 6 \]
\[ 2x^2 = 6 - y \]
\[ x^2 = \frac{6 - y}{2} \]
5. This is in the form \( x^2 = -\frac{1}{2}y + 3 \), which is a parabola.

**Conclusion for Option B:** This represents a parabola.

### Step 3: Analyze Option C
**Given:**
- \( \sqrt{x} = \tan t \)
- \( \sqrt{y} = \sec t \)

**Solution:**
1. Squaring both sides gives:
\[ x = \tan^2 t \]
\[ y = \sec^2 t \]
2. Using the identity \( \sec^2 t - \tan^2 t = 1 \):
\[ y - x = 1 \]
3. This represents a straight line.

**Conclusion for Option C:** Not a parabola.

### Step 4: Analyze Option D
**Given:**
- \( x = \sqrt{1 - \sin t} \)
- \( y = \sqrt{\sin \frac{t}{2} + \cos \frac{t}{2}} \)

**Solution:**
1. Squaring both equations:
\[ x^2 = 1 - \sin t \]
\[ y^2 = \sin \frac{t}{2} + \cos \frac{t}{2} \]
2. Using the identity \( \sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2} \):
\[ x^2 + \sin t = 1 \]
3. Rearranging gives:
\[ x^2 + 2 \sin \frac{t}{2} \cos \frac{t}{2} = 1 \]
4. This does not simplify to a parabolic equation.

**Conclusion for Option D:** Not a parabola.

### Final Conclusion:
The only option that represents a parabolic profile is **Option B**.