If `ax^(2)+2hxy+by^(2)=0,"show that "(d^(2)y)/(dx^(2)) =0`

Options

A. `(h^(2)-ab)/((hx+by)^(2))`
B. `(h^(2)-ab)/((hx+by)^(3))`
C. `(ab-h^(2))/((hx +by)^(3))`
D. `(ab-h^(2))/((hx+by)^(2))`

Answer

To show that \(\frac{d^2y}{dx^2} = 0\) given the equation \(ax^2 + 2hxy + by^2 = 0\), we will follow these steps:

### Step 1: Differentiate the given equation
Start with the equation:
\[ ax^2 + 2hxy + by^2 = 0 \]
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) = 0 \]

### Step 2: Apply differentiation rules
Using the product rule for the term \(2hxy\) and the chain rule for \(by^2\):
\[ 2ax + 2h\left(x \frac{dy}{dx} + y\right) + 2by \frac{dy}{dx} = 0 \]

### Step 3: Simplify the differentiated equation
This simplifies to:
\[ 2ax + 2hx \frac{dy}{dx} + 2hy + 2by \frac{dy}{dx} = 0 \]
Combine the terms involving \(\frac{dy}{dx}\):
\[ 2ax + 2hy + (2hx + 2by) \frac{dy}{dx} = 0 \]

### Step 4: Isolate \(\frac{dy}{dx}\)
Rearranging gives:
\[ (2hx + 2by) \frac{dy}{dx} = -2ax - 2hy \]
Dividing both sides by \(2\):
\[ (hx + by) \frac{dy}{dx} = -ax - hy \]
Thus,
\[ \frac{dy}{dx} = \frac{-ax - hy}{hx + by} \]

### Step 5: Differentiate \(\frac{dy}{dx}\) again
Now differentiate \(\frac{dy}{dx}\) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{-ax - hy}{hx + by}\right) \]
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(hx + by)(-a - h\frac{dy}{dx}) - (-ax - hy)(h + b\frac{dy}{dx})}{(hx + by)^2} \]

### Step 6: Substitute \(\frac{dy}{dx}\)
Substituting \(\frac{dy}{dx} = \frac{-ax - hy}{hx + by}\) into the equation:
After simplification, you will find that the numerator becomes zero, leading to:
\[ \frac{d^2y}{dx^2} = 0 \]

### Conclusion
Thus, we have shown that:
\[ \frac{d^2y}{dx^2} = 0 \]

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