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 be \((x_1,y_1)\).
Since the point lies on the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] we have \[ \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1. \]

Step 2: Write the chord of contact.
For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=2, \] the chord of contact from \((x_1,y_1)\) is \[ \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=2. \]

Step 3: Find intersections with the asymptotes.
The asymptotes of both hyperbolas are \[ \frac{x}{a}+\frac{y}{b}=0 \] and \[ \frac{x}{a}-\frac{y}{b}=0. \] Let \[ X=\frac{x}{a},\qquad Y=\frac{y}{b}. \] Then the asymptotes become \[ X+Y=0 \] and \[ X-Y=0. \] Also, put \[ u=\frac{x_1}{a},\qquad v=\frac{y_1}{b}. \] Since \((x_1,y_1)\) lies on the first hyperbola, \[ u^2-v^2=1. \] The chord of contact becomes \[ uX-vY=2. \]

Step 4: Find the intercepts made by the chord on the asymptotes.
On the asymptote \[ Y=X, \] we get \[ uX-vX=2. \] \[ X(u-v)=2. \] So, \[ X=\frac{2}{u-v}. \] Hence one point is \[ \left(\frac{2}{u-v},\frac{2}{u-v}\right). \] On the asymptote \[ Y=-X, \] we get \[ uX+vX=2. \] \[ X(u+v)=2. \] So, \[ X=\frac{2}{u+v}. \] Hence the other point is \[ \left(\frac{2}{u+v},-\frac{2}{u+v}\right). \]

Step 5: Use \(u^2-v^2=1\).
Since \[ u^2-v^2=(u-v)(u+v)=1, \] the above points simplify naturally while calculating area.
The triangle formed by the chord of contact and the two asymptotes has vertices at the origin and the two points found above.
In the \(XY\)-plane, its area is \[ \frac12 \left| \begin{vmatrix} \frac{2}{u-v} & \frac{2}{u-v} \frac{2}{u+v} & -\frac{2}{u+v} \end{vmatrix} \right|. \] \[ = \frac12 \left| -\frac{4}{(u-v)(u+v)} -\frac{4}{(u-v)(u+v)} \right|. \] \[ = \frac12 \left| -\frac{8}{u^2-v^2} \right|. \] Since \[ u^2-v^2=1, \] area in the \(XY\)-plane is \[ 4. \]

Step 6: Convert area back to the \(xy\)-plane.
Since \[ x=aX,\qquad y=bY, \] the area scale factor is \[ ab. \] Therefore, required area is \[ 4ab. \]

Step 7: Final conclusion.
Hence, \[ \boxed{4ab} \]