If `x=asinalphaandy=bcosalpha`, then `b^(2)x^(2)+a^(2)y^(2)=?`

Options

A. 1
B. 2
C. `a^(2)b^(2)`
D. `a^(2)+b^(2)`

Answer

To solve the problem, we need to find the value of the expression \( b^2 x^2 + a^2 y^2 \) given that \( x = a \sin \alpha \) and \( y = b \cos \alpha \).

### Step-by-step Solution:

1. **Substitute the values of \( x \) and \( y \)**:
\[ x = a \sin \alpha \quad \text{and} \quad y = b \cos \alpha \]

2. **Calculate \( x^2 \)**:
\[ x^2 = (a \sin \alpha)^2 = a^2 \sin^2 \alpha \]

3. **Calculate \( y^2 \)**:
\[ y^2 = (b \cos \alpha)^2 = b^2 \cos^2 \alpha \]

4. **Substitute \( x^2 \) and \( y^2 \) into the expression \( b^2 x^2 + a^2 y^2 \)**:
\[ b^2 x^2 = b^2 (a^2 \sin^2 \alpha) = a^2 b^2 \sin^2 \alpha \]
\[ a^2 y^2 = a^2 (b^2 \cos^2 \alpha) = a^2 b^2 \cos^2 \alpha \]

5. **Combine the two results**:
\[ b^2 x^2 + a^2 y^2 = a^2 b^2 \sin^2 \alpha + a^2 b^2 \cos^2 \alpha \]

6. **Factor out the common term \( a^2 b^2 \)**:
\[ b^2 x^2 + a^2 y^2 = a^2 b^2 (\sin^2 \alpha + \cos^2 \alpha) \]

7. **Use the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \)**:
\[ b^2 x^2 + a^2 y^2 = a^2 b^2 \cdot 1 = a^2 b^2 \]

### Final Answer:
\[ b^2 x^2 + a^2 y^2 = a^2 b^2 \]