Direction: Consider the following for the next 03 (three) items: If p = X cos θ - Y sin θ, q = X sin θ + Y cos θ and p 2 + 4pq + q 2 = AX 2 + BY 2, 0 ≤ θ ≤ π/2
What is the value of B?
-1
Understanding the Problem: Finding the Value of B
The question provides two linear equations relating variables \( p \) and \( q \) to \( X \) and \( Y \) using trigonometric functions of an angle \( \theta \). It also gives a quadratic equation relating \( p \) and \( q \) in the form \( p^2 + 4pq + q^2 \), which is said to be equal to a quadratic form in \( X \) and \( Y \), \( AX^2 + BY^2 \). We are asked to find the specific value of the constant \( B \) in this equation.
The key to solving this problem is to substitute the given expressions for \( p \) and \( q \) in terms of \( X \) and \( Y \) into the equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \). By expanding and simplifying the left side of the equation, we can express it in the form \( (\text{coefficient of } X^2)X^2 + (\text{coefficient of } Y^2)Y^2 + (\text{coefficient of } XY)XY \). Since the right side, \( AX^2 + BY^2 \), has no \( XY \) term, the coefficient of \( XY \) on the left side must be zero. This condition will help us determine the value of \( \theta \). Once \( \theta \) is known, we can find the coefficients of \( X^2 \) and \( Y^2 \) on the left side and equate the coefficient of \( Y^2 \) to \( B \) to find its value.
Step-by-Step Solution
We are given:
\( p = X \cos \theta - Y \sin \theta \)
\( q = X \sin \theta + Y \cos \theta \)
\( p^2 + 4pq + q^2 = AX^2 + BY^2 \)
Let's calculate \( p^2 \), \( q^2 \), and \( pq \).
Calculate \( p^2 \):
\[ p^2 = (X \cos \theta - Y \sin \theta)^2 \] \[ p^2 = (X \cos \theta)^2 - 2(X \cos \theta)(Y \sin \theta) + (Y \sin \theta)^2 \] \[ p^2 = X^2 \cos^2 \theta - 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta \]Calculate \( q^2 \):
\[ q^2 = (X \sin \theta + Y \cos \theta)^2 \] \[ q^2 = (X \sin \theta)^2 + 2(X \sin \theta)(Y \cos \theta) + (Y \cos \theta)^2 \] \[ q^2 = X^2 \sin^2 \theta + 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta \]Calculate \( pq \):
\[ pq = (X \cos \theta - Y \sin \theta)(X \sin \theta + Y \cos \theta) \] \[ pq = X \cos \theta (X \sin \theta + Y \cos \theta) - Y \sin \theta (X \sin \theta + Y \cos \theta) \] \[ pq = X^2 \cos \theta \sin \theta + XY \cos^2 \theta - XY \sin^2 \theta - Y^2 \sin \theta \cos \theta \] \[ pq = (X^2 - Y^2) \sin \theta \cos \theta + XY (\cos^2 \theta - \sin^2 \theta) \]Now, substitute these into the equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \):
\[ (X^2 \cos^2 \theta - 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta) \] \[ + 4[(X^2 - Y^2) \sin \theta \cos \theta + XY (\cos^2 \theta - \sin^2 \theta)] \] \[ + (X^2 \sin^2 \theta + 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta) \] \[ = AX^2 + BY^2 \]Group the terms by \( X^2 \), \( Y^2 \), and \( XY \):
Coefficient of \( X^2 \):
\[ \cos^2 \theta + 4 \sin \theta \cos \theta + \sin^2 \theta \] \[ = (\cos^2 \theta + \sin^2 \theta) + 4 \sin \theta \cos \theta \] \[ = 1 + 4 \sin \theta \cos \theta \]Coefficient of \( Y^2 \):
\[ \sin^2 \theta - 4 \sin \theta \cos \theta + \cos^2 \theta \] \[ = (\sin^2 \theta + \cos^2 \theta) - 4 \sin \theta \cos \theta \] \[ = 1 - 4 \sin \theta \cos \theta \]Coefficient of \( XY \):
\[ -2 \cos \theta \sin \theta + 4 (\cos^2 \theta - \sin^2 \theta) + 2 \sin \theta \cos \theta \] \[ = 4 (\cos^2 \theta - \sin^2 \theta) \]So, the expanded equation is:
\[ X^2 (1 + 4 \sin \theta \cos \theta) + Y^2 (1 - 4 \sin \theta \cos \theta) + XY (4 (\cos^2 \theta - \sin^2 \theta)) = AX^2 + BY^2 \]By comparing the coefficients of \( X^2 \), \( Y^2 \), and \( XY \) on both sides of the equation:
Coefficient of \( X^2 \): \( A = 1 + 4 \sin \theta \cos \theta \)
Coefficient of \( Y^2 \): \( B = 1 - 4 \sin \theta \cos \theta \)
Coefficient of \( XY \): \( 4 (\cos^2 \theta - \sin^2 \theta) = 0 \)
From the coefficient of \( XY \), we have \( \cos^2 \theta - \sin^2 \theta = 0 \). This is the double angle identity for cosine, \( \cos(2\theta) = 0 \). Alternatively, \( \cos^2 \theta = \sin^2 \theta \). Given the range \( 0 \le \theta \le \pi/2 \), where both \( \sin \theta \) and \( \cos \theta \) are non-negative, we must have \( \cos \theta = \sin \theta \). This condition is satisfied when \( \theta = \pi/4 \).
Now substitute \( \theta = \pi/4 \) into the expression for \( B \):
\[ \sin(\pi/4) = \frac{1}{\sqrt{2}} \] \[ \cos(\pi/4) = \frac{1}{\sqrt{2}} \] \[ \sin \theta \cos \theta = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2} \] \[ B = 1 - 4 \sin \theta \cos \theta \] \[ B = 1 - 4 \left(\frac{1}{2}\right) \] \[ B = 1 - 2 \] \[ B = -1 \]The value of \( B \) is \( -1 \).
Summary of Coefficients
| Term | Coefficient on Left Side | Coefficient on Right Side |
|---|---|---|
| \( X^2 \) | \( 1 + 4 \sin \theta \cos \theta \) | \( A \) |
| \( Y^2 \) | \( 1 - 4 \sin \theta \cos \theta \) | \( B \) |
| \( XY \) | \( 4 (\cos^2 \theta - \sin^2 \theta) \) | \( 0 \) |
Equating the coefficient of \( XY \) to zero leads to \( \theta = \pi/4 \). Substituting this value into the coefficient of \( Y^2 \) gives \( B = 1 - 4 \sin(\pi/4) \cos(\pi/4) = 1 - 4(1/\sqrt{2})(1/\sqrt{2}) = 1 - 4(1/2) = 1 - 2 = -1 \).
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Linear Transformation | Equations relating variables (p, q) to (X, Y) linearly. | Used to substitute p and q into the quadratic equation. |
| Quadratic Form | An expression like \( AX^2 + BY^2 + CXY \). | The given equation \( p^2 + 4pq + q^2 \) transforms into this form in terms of X and Y. |
| Trigonometric Identities | Equations involving trigonometric functions (e.g., \( \sin^2\theta + \cos^2\theta = 1 \), \( \cos^2\theta - \sin^2\theta = \cos(2\theta) \)). | Essential for simplifying the expressions after substitution and for determining the value of \( \theta \). |
| Coefficient Comparison | Equating coefficients of like terms on both sides of an identity or equation. | Used to find the values of A and B after expressing the left side in terms of \( X^2, Y^2, XY \). |
Additional Information: Orthogonal Transformations
The transformation \( p = X \cos \theta - Y \sin \theta \) and \( q = X \sin \theta + Y \cos \theta \) is a common linear transformation, specifically a rotation of coordinates in a 2D plane if \( \theta \) is interpreted as the angle of rotation. When \( \theta = \pi/4 \), this corresponds to a rotation by 45 degrees. Such transformations are often called orthogonal transformations because they preserve distances and angles. In matrix form, this transformation is:
\[ \begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} \]The matrix is a rotation matrix, and its transpose is equal to its inverse, which is a property of orthogonal matrices.
The expression \( p^2 + q^2 \) is invariant under this type of transformation, meaning \( p^2 + q^2 = X^2 + Y^2 \). Let's verify for \( \theta = \pi/4 \):
\[ p = X \frac{1}{\sqrt{2}} - Y \frac{1}{\sqrt{2}} \] \[ q = X \frac{1}{\sqrt{2}} + Y \frac{1}{\sqrt{2}} \] \[ p^2 = \frac{1}{2}(X-Y)^2 = \frac{1}{2}(X^2 - 2XY + Y^2) \] \[ q^2 = \frac{1}{2}(X+Y)^2 = \frac{1}{2}(X^2 + 2XY + Y^2) \] \[ p^2 + q^2 = \frac{1}{2}(X^2 - 2XY + Y^2 + X^2 + 2XY + Y^2) = \frac{1}{2}(2X^2 + 2Y^2) = X^2 + Y^2 \]This confirms the invariance of \( p^2 + q^2 \).
In our specific problem, the equation involves \( p^2 + 4pq + q^2 \), which is not a simple sum of squares, and therefore its transformation into \( AX^2 + BY^2 \) involves more than just \( X^2 + Y^2 \) terms unless the \( XY \) term vanishes, which dictated the value of \( \theta \).
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