Question

Let O be the origin, and P and Q be two points on the rectangular hyperbola \( xy = 12 \) such that the midpoint of the line segment PQ is \( \left( \frac{1}{2}, -\frac{1}{2} \right) \). Then the area of the triangle OPQ equals:

• A. \( \frac{3}{2} \)
• B. \( \frac{5}{2} \)
• C. \( \frac{7}{2} \)
• D. \( \frac{9}{2} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a conic section, the equation of a chord with a given midpoint \((x_1, y_1)\) is given by \(T = S_1\).
After finding the equation of the line \(PQ\), we can solve it simultaneously with the curve \(xy = 12\) to find the exact coordinates of points \(P\) and \(Q\).
Finally, we use the determinant method to find the area of triangle \(OPQ\).
Step 2: Key Formula or Approach:
The equation of the hyperbola is \(S \equiv xy - 12 = 0\).
The midpoint is \((x_1, y_1) = \left(\frac{1}{2}, -\frac{1}{2}\right)\).
The chord formula is \(T = S_1\), which translates to \(\frac{xy_1 + yx_1}{2} - 12 = x_1y_1 - 12\).
Step 3: Detailed Explanation:
Substitute the coordinates of the midpoint into the chord formula.
\[ \frac{x(-1/2) + y(1/2)}{2} = (1/2)(-1/2) \] \[ \frac{-x + y}{4} = \frac{-1}{4} \] Multiply by 4 to get the equation of the line \(PQ\).
\[ -x + y = -1 \implies y = x - 1 \] Now, substitute \(y = x - 1\) into the hyperbola equation \(xy = 12\) to find points \(P\) and \(Q\).
\[ x(x - 1) = 12 \] \[ x^2 - x - 12 = 0 \] Factor the quadratic equation.
\[ (x - 4)(x + 3) = 0 \] This gives the \(x\)-coordinates of \(P\) and \(Q\) as \(x = 4\) and \(x = -3\).
Using \(y = x - 1\), the corresponding \(y\)-coordinates are \(y = 3\) and \(y = -4\).
So, the points are \(P(4, 3)\) and \(Q(-3, -4)\).
The area of triangle \(OPQ\) with the origin \(O(0,0)\) is given by the formula \(\frac{1}{2}|x_1y_2 - x_2y_1|\).
\[ \text{Area} = \frac{1}{2} |(4)(-4) - (-3)(3)| \] \[ \text{Area} = \frac{1}{2} |-16 + 9| \] \[ \text{Area} = \frac{1}{2} |-7| = \frac{7}{2} \] Step 4: Final Answer:
The area of the triangle \(OPQ\) equals \(\frac{7}{2}\).