If `y ^(2) =4ax,` then `(d^(2) y)/(dx ^(2))=(ka ^(2))/( y ^(3)), where k ^(2)=`

Answer

To solve the problem, we need to find the value of \( k^2 \) given the equation \( y^2 = 4ax \) and the relationship \( \frac{d^2y}{dx^2} = \frac{ka^2}{y^3} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ y^2 = 4ax \]

2. **Differentiate both sides with respect to \( x \):**
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 4a \]

3. **Solve for \( \frac{dy}{dx} \):**
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

4. **Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):**
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2a}{y}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = -\frac{2a \cdot \frac{dy}{dx}}{y^2} \]

5. **Substitute \( \frac{dy}{dx} \) into the equation:**
\[ \frac{d^2y}{dx^2} = -\frac{2a \cdot \frac{2a}{y}}{y^2} = -\frac{4a^2}{y^3} \]

6. **Relate this to the given equation:**
According to the problem, we have:
\[ \frac{d^2y}{dx^2} = \frac{ka^2}{y^3} \]
Setting the two expressions equal:
\[ -\frac{4a^2}{y^3} = \frac{ka^2}{y^3} \]

7. **Cancel \( a^2/y^3 \) (assuming \( a \neq 0 \) and \( y \neq 0 \)):**
\[ -4 = k \]

8. **Find \( k^2 \):**
\[ k^2 = (-4)^2 = 16 \]

### Final Answer:
\[ k^2 = 16 \]