91Ó°ÊÓ

The general equation of a hyperbola centered at the origin with major axis along the x-axis is \[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\] where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. The eccentricity of the hyperbola is given by \[e=\frac{\sqrt{a^2+b^2}}{a}.\] Now, we have two hyperbolas, so let's denote their parameters by \((a_1,b_1,e_1)\) and \((a_2,b_2,e_2)\), with the first hyperbola having its major axis along the x-axis, and the second along the y-axis.

The asymptotes of the hyperbolas are the lines that follow the equation \[\frac{x^2}{a^2}-\frac{y^2}{b^2}=0.\] For the two given hyperbolas, their equations will be \[\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=0\] and \[\frac{y^2}{a_2^2}-\frac{x^2}{b_2^2}=0.\]

Since the hyperbolas share a set of asymptotes, their asymptote equations must be equal. Cross multiply and simplify: \[x^2b_2^2=a_1^2y^2 \quad \text{and} \quad x^2a_2^2=b_1^2y^2.\] Now let's take the ratio of these two equations: \[\frac{x^2b_2^2}{a_1^2y^2} = \frac{x^2a_2^2}{b_1^2y^2}.\] After cancelling x^2 and y^2, we get the following relationship: \[\frac{b_2^2}{a_1^2}=\frac{a_2^2}{b_1^2}.\]

Firstly, let's denote our eccentricities E and e just as in the exercise. Let \(E = e_1\) and \(e = e_2\). We recall the relationship between the eccentricity and the semi-axes of a hyperbola: \[E=\frac{\sqrt{a_1^2+b_1^2}}{a_1}, \quad e=\frac{\sqrt{a_2^2+b_2^2}}{a_2}.\] Squaring these relationships, we obtain: \[E^2=\frac{a_1^2+b_1^2}{a_1^2}, \quad e^2=\frac{a_2^2+b_2^2}{a_2^2}.\] Now, we can rewrite these equations as \[b_1^2=a_1^2(E^2-1), \quad b_2^2=a_2^2(e^2-1).\] Substituting these into the relationship found in step 3: \[\frac{a_2^2(e^2-1)}{a_1^2}=\frac{a_1^2(E^2-1)}{a_2^2}.\] Now, we simply cancel out \(a_1^2a_2^2\) and rearrange to form the expression we are asked to prove: \[e^2E^2-1=(E^2-1)(e^2-1).\] \[1 - e^2E^2 = (1 - E^2)(1 - e^2).\] \[1=e^2-e^2E^2+E^2-E^2e^2.\] \[1=E^{-2}-1+e^{-2}.\] Finally, we get the desired expression: \[e^{-2}+E^{-2}=1.\]