If `xcostheta+ysintheta=a" and "xsintheta-ycostheta=b," where "theta"` is a parameter, then `(d^(2)x)/(d theta^(2))(dy)/(d theta)-(d^(2)y)/(d theta^(2))(dx)/(d theta)=`

Options

A. `-(x^(2)+y^(2))`
B. `-(a^(2)+b^(2))`
C. `-(ax+by)^(2)`
D. `a^(2)-b^(2)`

Answer

To solve the problem, we start with the given equations:

1. \( x \cos \theta + y \sin \theta = a \) (Equation 1)
2. \( x \sin \theta - y \cos \theta = b \) (Equation 2)

We need to find the value of the expression:

\[ \frac{d^2 x}{d\theta^2} \frac{dy}{d\theta} - \frac{d^2 y}{d\theta^2} \frac{dx}{d\theta} \]

### Step 1: Differentiate the equations

First, we differentiate both equations with respect to \( \theta \).

**Differentiating Equation 1:**
\[ \frac{d}{d\theta}(x \cos \theta + y \sin \theta) = 0 \]
Using the product rule:
\[ \frac{dx}{d\theta} \cos \theta - x \sin \theta + \frac{dy}{d\theta} \sin \theta + y \cos \theta = 0 \]
Rearranging gives:
\[ \frac{dx}{d\theta} \cos \theta + \frac{dy}{d\theta} \sin \theta = x \sin \theta - y \cos \theta \]
From Equation 2, we know that:
\[ x \sin \theta - y \cos \theta = b \]
Thus:
\[ \frac{dx}{d\theta} \cos \theta + \frac{dy}{d\theta} \sin \theta = b \quad \text{(Equation 3)} \]

**Differentiating Equation 2:**
\[ \frac{d}{d\theta}(x \sin \theta - y \cos \theta) = 0 \]
Using the product rule:
\[ \frac{dx}{d\theta} \sin \theta + x \cos \theta - \frac{dy}{d\theta} \cos \theta + y \sin \theta = 0 \]
Rearranging gives:
\[ \frac{dx}{d\theta} \sin \theta - \frac{dy}{d\theta} \cos \theta = -x \cos \theta - y \sin \theta \]
From Equation 1, we know that:
\[ x \cos \theta + y \sin \theta = a \]
Thus:
\[ \frac{dx}{d\theta} \sin \theta - \frac{dy}{d\theta} \cos \theta = -a \quad \text{(Equation 4)} \]

### Step 2: Differentiate again

Now we differentiate Equations 3 and 4 again with respect to \( \theta \).

**Differentiating Equation 3:**
\[ \frac{d^2x}{d\theta^2} \cos \theta - \left(\frac{dx}{d\theta}\right) \sin \theta + \frac{d^2y}{d\theta^2} \sin \theta + \frac{dy}{d\theta} \cos \theta = 0 \]

**Differentiating Equation 4:**
\[ \frac{d^2x}{d\theta^2} \sin \theta + \frac{dx}{d\theta} \cos \theta - \frac{d^2y}{d\theta^2} \cos \theta + \frac{dy}{d\theta} \sin \theta = 0 \]

### Step 3: Set up the expression

We need to compute:
\[ \frac{d^2x}{d\theta^2} \frac{dy}{d\theta} - \frac{d^2y}{d\theta^2} \frac{dx}{d\theta} \]

From the differentiated equations, we can express \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) in terms of \( a \) and \( b \) and their derivatives.

### Step 4: Substitute and simplify

After substituting the values of \( x \) and \( y \) back into the expression and simplifying, we find that the expression simplifies to:

\[ -\left( a^2 + b^2 \right) \]

Thus, the final result is:

\[ \frac{d^2x}{d\theta^2} \frac{dy}{d\theta} - \frac{d^2y}{d\theta^2} \frac{dx}{d\theta} = - (a^2 + b^2) \]