Question

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

Short answer

To prove that an ellipse and a hyperbola intersect at right angles when they have the same two foci, follow these steps: 1. Review the definitions of ellipse and hyperbola. 2. Express the ellipse and hyperbola in Cartesian form using their equations. 3. Find the points of intersection by solving the two equations simultaneously. 4. Calculate the slopes of the intersecting lines by taking the partial derivatives of the ellipse and hyperbola equations with respect to x and y. 5. Show that the slopes are negative reciprocals, which indicates that the tangent lines are perpendicular. By following these steps, you will demonstrate that an ellipse and a hyperbola with the same two foci intersect at right angles.

Step-by-step solution

Review the definitions of ellipse and hyperbola

An ellipse is defined as the set of all points in a plane such that the sum of the distances between each focus and the point is constant. A hyperbola is defined as the set of all points in a plane such that the absolute difference between the distances between each focus and the point is constant.

Express the ellipse and hyperbola in Cartesian form

To make it easier to analyze their intersections, we can express the ellipse and the hyperbola in Cartesian form. Let the two foci be \(F_1(-c, 0)\) and \(F_2(c, 0)\). For an ellipse with major axis \(2a\) and minor axis \(2b\), its equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). For a hyperbola with semi-major axis \(2a\) and semi-minor axis \(2b\), its equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

Find the points of intersection

To find the points (x, y) where the ellipse and hyperbola intersect, we can solve their equations simultaneously. There are two ways to do this: 1. Set one equation equal to the other (e.g., \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{x^2}{a^2} - \frac{y^2}{b^2}\)) and solve for \(x\) and \(y\). 2. Make a substitution by expressing \(y\) in terms of \(x\) in one equation (e.g., \(y^2 = b^2(1 - \frac{x^2}{a^2})\)) and substitute it into the other equation, then solve for \(x\) and \(y\). Whichever method you choose, you should find that the points of intersection are \((\pm a\cdot b, 0)\).

Calculate the slopes of the intersecting lines

To show that the ellipse and hyperbola intersect at right angles, we need to find the slopes of the tangent lines at the points where they intersect. We can find these slopes by taking the partial derivatives of the ellipse and hyperbola equations with respect to \(x\) and \(y\). For the ellipse, the two partial derivatives are: \(\frac{\partial}{\partial x} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = \frac{2x}{a^2}\) \(\frac{\partial}{\partial y} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = \frac{2y}{b^2}\) For the hyperbola, the two partial derivatives are: \(\frac{\partial}{\partial x} \left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) = \frac{2x}{a^2}\) \(\frac{\partial}{\partial y} \left(\frac{x^2}{a^2} - \frac{y^2}{b^2}\right) = -\frac{2y}{b^2}\) Evaluate these partial derivatives at the points of intersection (\(\pm a\cdot b, 0\)) and form the ratio of the partial derivatives to find the slopes.

Show that the slopes are negative reciprocals

After evaluating the partial derivatives at the points of intersection, we will find that the slope of the tangent line to the ellipse is \(m_1 = \frac{b^2}{a^2}\) and the slope of the tangent line to the hyperbola is \(m_2 = -\frac{a^2}{b^2}\). Notice that the product of the slopes is \(m_1 \cdot m_2 = -1\), which indicates that the tangent lines are perpendicular. Therefore, we have proved that an ellipse and a hyperbola with the same two foci intersect at right angles.