What is the differential equation to family of parabola having their vertices at the origin and foci on the X-axis ?

Options

A. y = 2x y'
B. x = 2y y'
C. xy = y'
D. x = yy'

Answer

To find the differential equation for the family of parabolas with vertices at the origin and foci on the X-axis, we start with the standard equation of such parabolas.

### Step-by-Step Solution:

1. **Identify the Standard Equation of the Parabola:**
The standard form of a parabola with its vertex at the origin and focus on the X-axis is given by:
\[ y^2 = 4ax \]
where \( a \) is the distance from the vertex to the focus.

2. **Differentiate the Equation:**
We differentiate both sides of the equation \( y^2 = 4ax \) with respect to \( x \). Using implicit differentiation:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \]
This gives us:
\[ 2y \frac{dy}{dx} = 4a \]
We can denote \( \frac{dy}{dx} \) as \( y' \):
\[ 2y y' = 4a \]

3. **Solve for \( a \):**
Rearranging the equation to solve for \( a \):
\[ a = \frac{2y y'}{4} = \frac{y y'}{2} \]

4. **Substitute \( a \) back into the original equation:**
Now, substitute the expression for \( a \) back into the original parabola equation:
\[ y^2 = 4\left(\frac{y y'}{2}\right)x \]
Simplifying this gives:
\[ y^2 = 2xy y' \]

5. **Rearranging the Equation:**
We can rearrange the equation to express it in a standard form:
\[ y = 2xy' \]

### Final Differential Equation:
The differential equation for the family of parabolas having their vertices at the origin and foci on the X-axis is:
\[ y = 2xy' \]