Find the differential equation of all parabolas whose axes are parallel to the x-axis an having latus rectum a.

Explanation

To find the differential equation of all parabolas whose axes are parallel to the x-axis and having latus rectum \( a \), we can follow these steps:

### Step 1: Write the equation of the parabola
The standard form of the equation of a parabola with its axis parallel to the x-axis and having latus rectum \( a \) is given by:

\[ (y - \beta)^2 = a(x - \alpha) \]

where \( \alpha \) and \( \beta \) are constants representing the vertex of the parabola.

### Step 2: Differentiate the equation
We differentiate the equation with respect to \( x \):

\[ \frac{d}{dx}[(y - \beta)^2] = \frac{d}{dx}[a(x - \alpha)] \]

Using the chain rule on the left side, we get:

\[ 2(y - \beta) \frac{dy}{dx} = a \]

This can be rewritten as:

\[ 2(y - \beta) \frac{dy}{dx} - a = 0 \tag{1} \]

### Step 3: Differentiate again
Now we differentiate equation (1) again with respect to \( x \):

Using the product rule and chain rule, we have:

\[ \frac{d}{dx}[2(y - \beta) \frac{dy}{dx}] = 0 \]

This gives us:

\[ 2\left(\frac{dy}{dx}(y - \beta) + (y - \beta) \frac{d^2y}{dx^2}\right) = 0 \]

Simplifying, we get:

\[ (y - \beta) \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0 \]

### Step 4: Substitute from the first derivative
From equation (1), we know that:

\[ 2(y - \beta) \frac{dy}{dx} = a \]

Thus, we can substitute \( 2(y - \beta) \) into our second derivative equation:

\[ \frac{d^2y}{dx^2} = -\frac{2}{a} \left(\frac{dy}{dx}\right)^2 \]

### Step 5: Final form of the differential equation
Rearranging gives us the final form of the differential equation:

\[ a \frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^3 = 0 \]

This is the required differential equation of all parabolas whose axes are parallel to the x-axis and having latus rectum \( a \).

---