Question

From any point on the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] tangents are drawn to the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=2. \] The area of the figure formed by the chord of contact of that point and the asymptotes is:

• A. \(\frac{ab}{2}\)
• B. \(ab\)
• C. \(2ab\)
• D. \(4ab\)

Hint:

For conics with scaled variables, use \(X=\frac{x}{a}\) and \(Y=\frac{y}{b}\). Area in the original \(xy\)-plane becomes \(ab\) times the area in the transformed \(XY\)-plane.

Answer 1

The Correct Option is D

Step 1: Let the external point lie on the first hyperbola.
The point (x₁, y₁) satisfies x₁²/a² - y₁²/b² = 1. Define u = x₁/a and v = y₁/b, giving u² - v² = 1.

Step 2: Write the chord of contact to the second hyperbola.
For x²/a² - y²/b² = 2, the chord of contact from (x₁, y₁) is xx₁/a² - yy₁/b² = 2, which becomes uX - vY = 2 in terms of X = x/a and Y = y/b.

Step 3: Find intersections with the asymptotes.
The asymptotes are Y = X and Y = -X. On Y = X: uX - vX = 2 → X = 2/(u - v), giving point (2/(u-v), 2/(u-v)). On Y = -X: uX + vX = 2 → X = 2/(u + v), giving point (2/(u+v), -2/(u+v)).

Step 4: Compute the triangle area in the XY-plane.
Vertices are the origin and the two intersection points. Area = ½|det| = ½|(2/(u-v))(-2/(u+v)) - (2/(u-v))(2/(u+v))| = 4/|u² - v²| = 4 (since u² - v² = 1).

Step 5: Scale back to the xy-plane.
The area scales by ab, giving 4ab.

Step 6: Final conclusion.
The area of the triangle is 4ab.