Understanding the Problem

The problem provides us with a relationship between two sets of variables, \((p, q)\) and \((X, Y)\), defined by trigonometric expressions involving an angle \(\theta\). We are also given a quadratic equation relating \(p\) and \(q\), which is stated to be equivalent to a quadratic equation in \(X\) and \(Y\).

The given equations are:

\( p = X \cos \theta - Y \sin \theta \)

\( q = X \sin \theta + Y \cos \theta \)

\( p^2 + 4pq + q^2 = AX^2 + BY^2 \)

We are also given that \( 0 \le \theta \le \pi/2 \). Our goal is to find the value of the coefficient \(A\) in the equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \).

Solving the Equation by Substitution

To find the values of \(A\) and \(B\), we need to substitute the expressions for \(p\) and \(q\) in terms of \(X\), \(Y\), and \(\theta\) into the equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \). Let's first calculate \(p^2\), \(q^2\), and \(pq\).

Calculating \(p^2\) and \(q^2\)

We square the expressions for \(p\) and \(q\):

\[ p^2 = (X \cos \theta - Y \sin \theta)^2 = X^2 \cos^2 \theta - 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta \]

\[ q^2 = (X \sin \theta + Y \cos \theta)^2 = X^2 \sin^2 \theta + 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta \]

Adding \(p^2\) and \(q^2\):

\[ p^2 + q^2 = (X^2 \cos^2 \theta - 2XY \cos \theta \sin \theta + Y^2 \sin^2 \theta) + (X^2 \sin^2 \theta + 2XY \sin \theta \cos \theta + Y^2 \cos^2 \theta) \]

Group terms by \(X^2\) and \(Y^2\):

\[ p^2 + q^2 = X^2 (\cos^2 \theta + \sin^2 \theta) + Y^2 (\sin^2 \theta + \cos^2 \theta) + (-2XY \cos \theta \sin \theta + 2XY \sin \theta \cos \theta) \]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), the expression simplifies to:

\[ p^2 + q^2 = X^2 (1) + Y^2 (1) + 0 = X^2 + Y^2 \]

Calculating \(pq\)

Now, we calculate the product \(pq\):

\[ pq = (X \cos \theta - Y \sin \theta)(X \sin \theta + Y \cos \theta) \]

Expand the product:

\[ pq = X \cos \theta (X \sin \theta + Y \cos \theta) - Y \sin \theta (X \sin \theta + Y \cos \theta) \]

\[ pq = X^2 \cos \theta \sin \theta + XY \cos^2 \theta - XY \sin^2 \theta - Y^2 \sin \theta \cos \theta \]

Group terms by \(XY\) and \( (X^2 - Y^2) \):

\[ pq = (X^2 - Y^2) \sin \theta \cos \theta + XY (\cos^2 \theta - \sin^2 \theta) \]

Using the double angle identities \( \sin(2\theta) = 2 \sin \theta \cos \theta \) and \( \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \), we get:

\[ pq = \frac{1}{2} (X^2 - Y^2) \sin(2\theta) + XY \cos(2\theta) \]

Substituting into the Given Equation

Substitute the expressions for \(p^2+q^2\) and \(pq\) into the equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \):

\[ (X^2 + Y^2) + 4 \left[ \frac{1}{2} (X^2 - Y^2) \sin(2\theta) + XY \cos(2\theta) \right] = AX^2 + BY^2 \]

Distribute the 4:

\[ X^2 + Y^2 + 2 (X^2 - Y^2) \sin(2\theta) + 4XY \cos(2\theta) = AX^2 + BY^2 \]

\[ X^2 + Y^2 + 2X^2 \sin(2\theta) - 2Y^2 \sin(2\theta) + 4XY \cos(2\theta) = AX^2 + BY^2 \]

Group terms by \(X^2\), \(Y^2\), and \(XY\) on the left side:

\[ X^2 (1 + 2 \sin(2\theta)) + Y^2 (1 - 2 \sin(2\theta)) + XY (4 \cos(2\theta)) = AX^2 + BY^2 \]

Comparing Coefficients

We are given that the left side equals \(AX^2 + BY^2\). This means the coefficient of the \(XY\) term on the left side must be zero, because there is no \(XY\) term on the right side.

So, we must have:

\[ 4 \cos(2\theta) = 0 \implies \cos(2\theta) = 0 \]

We are given that \( 0 \le \theta \le \pi/2 \). This means \( 0 \le 2\theta \le \pi \). Within this range, the only value for \(2\theta\) for which \( \cos(2\theta) = 0 \) is \( 2\theta = \pi/2 \).

Therefore, \( \theta = \pi/4 \).

Calculating the Value of A

Now that we have found the value of \(\theta\), we can substitute it back into the equation we derived:

\[ X^2 (1 + 2 \sin(2\theta)) + Y^2 (1 - 2 \sin(2\theta)) + XY (4 \cos(2\theta)) = AX^2 + BY^2 \]

Substitute \( \theta = \pi/4 \):

\[ 2\theta = 2(\pi/4) = \pi/2 \]

\[ \sin(2\theta) = \sin(\pi/2) = 1 \]

\[ \cos(2\theta) = \cos(\pi/2) = 0 \]

Substituting these values:

\[ X^2 (1 + 2 (1)) + Y^2 (1 - 2 (1)) + XY (4 (0)) = AX^2 + BY^2 \]

\[ X^2 (1 + 2) + Y^2 (1 - 2) + 0 = AX^2 + BY^2 \]

\[ 3X^2 + (-1)Y^2 = AX^2 + BY^2 \]

\[ 3X^2 - Y^2 = AX^2 + BY^2 \]

Comparing the coefficients of \(X^2\) and \(Y^2\) on both sides:

Coefficient of \(X^2\): \( A = 3 \)

Coefficient of \(Y^2\): \( B = -1 \)

The question asks for the value of \(A\).

The value of \(A\) is 3.

Summary of the Solution Steps

Substitute the expressions for \(p\) and \(q\) into the given quadratic equation in terms of \(p\) and \(q\).

Simplify the resulting expression in terms of \(X\), \(Y\), and \(\theta\).

Compare the coefficients of \(X^2\), \(Y^2\), and \(XY\) with the target equation \(AX^2 + BY^2\).

Equate the coefficient of \(XY\) to zero to find the value of \(\theta\).

Substitute the found value of \(\theta\) back into the expression and identify the coefficient of \(X^2\) as \(A\).

Revision Table: Key Calculations

Additional Information: Coordinate Transformations

The initial equations \( p = X \cos \theta - Y \sin \theta \) and \( q = X \sin \theta + Y \cos \theta \) represent a rotation of the coordinate system. If we consider \((X, Y)\) as coordinates in one system, then \((p, q)\) are the coordinates of the same point in a system rotated by an angle \(\theta\). The expression \(p^2 + q^2 = X^2 + Y^2\) confirms this, as the squared distance from the origin is invariant under rotation.

The equation \( p^2 + 4pq + q^2 = AX^2 + BY^2 \) represents a quadratic form in the variables \(p\) and \(q\) being transformed into a quadratic form in \(X\) and \(Y\). By setting the coefficient of the mixed term (\(XY\)) to zero, we effectively find the angle of rotation \(\theta\) that eliminates the mixed product term (\(pq\)) in the original quadratic form \(p^2 + 4pq + q^2\), transforming it into a simpler form involving only squared terms \(X^2\) and \(Y^2\). This process is related to diagonalizing the matrix associated with the quadratic form.