Problem 95 - Calculus Early Transcendentals

Question

Show that an ellipse and a hyperbola that have the same two foci intersect at right angles.

Step-by-Step Solution

Answer

Answer: Yes, an ellipse and a hyperbola with the same foci intersect at right angles.

Step 1: Derive parametric equations of the ellipse and the hyperbola

Let the foci of both the ellipse and the hyperbola be F1 and F2, with distance 'c' between them. Ellipse equation with a horizontal major axis and center (h, k) is: (1) \[((x - h)^2)/a^2 + ((y - k)^2)/b^2 = 1\], where a is the semi-major axis, b is the semi-minor axis, and c = sqrt(a^2 - b^2), which is the distance between the center and the foci. Let's have a parameter 't_e' for the ellipse. Then we have: (2) \[x = h + a \cos{t_e}\] (3) \[y = k + b \sin{t_e}\] Now let's find the equations for a hyperbola, which has the form: (1') \[((x - h)^2)/a'^2 - ((y - k)^2)/b'^2 = 1\], where a' is the semi-major axis, b' is the semi-minor axis, and c' = sqrt(a'^2 + b'^2), which is the distance between the center and the foci. Let's have a parameter 't_h' for the hyperbola. Then we have: (2') \[x = h + a' \cosh{t_h} \] (3') \[y = k + b' \sinh{t_h}\]

Step 2: Find the derivatives to get tangent lines' slopes at intersection points

Let's find the derivatives of the ellipse equations: (4) \[dx/dt_e = -a \sin{t_e}\] (5) \[dy/dt_e = b \cos{t_e}\] The slope of the tangent to the ellipse, m_e = (dy/dt_e) / (dx/dt_e): (6) \[m_e = -\frac{b}{a}\cot{t_e}\] Now, let's find the derivatives of the hyperbola equations: (7) \[dx/dt_h = a' \sinh{t_h}\] (8) \[dy/dt_h = b' \cosh{t_h}\] The slope of the tangent to the hyperbola, m_h = (dy/dt_h) / (dx/dt_h): (9) \[m_h = \frac{b'}{a'}\coth{t_h}\]

Step 3: Prove that curves intersection is orthogonal

Now let's prove that the product of the slopes, m_e and m_h, at the intersection points is -1. (10) \[m_e * m_h = -\frac{b}{a}\cot{t_e} * \frac{b'}{a'}\coth{t_h} = -\frac{bb'}{aa'} \frac{\cosh{t_h}}{\sinh{t_h}} \frac{\sin{t_h}}{\cos{t_h}}\] Notice that the intersection points satisfy both the ellipse and hyperbola equations, so: (11) \[cosh^2{t_h} - sinh^2{t_h} = cos^2{t_e} + sin^2{t_e} = 1\] By using the identity (11), the product of the slopes at the intersection points becomes: (12) \[m_e * m_h = -\frac{bb'}{aa'}\] At the intersection points, a * a' = b * b' (by using ellipse and hyperbola focus equations). Thus: (13) \[m_e * m_h = -1\] We have shown that the product of the slopes at the intersection point is equal to -1, meaning the ellipse and the hyperbola intersect at right angles.