find a Cartesian equation for the conic section satisfying the given conditions. Sketch the curv

Find a Cartesian equation for the conic section satisfying...

Key Concepts

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. Each type has distinct geometric properties and unique standard equations, which are derived by analyzing their symmetry and defining characteristics.

Hyperbola

A hyperbola is a type of conic section defined as the set of all points in the plane where the absolute difference of their distances to two fixed points (known as the foci) is constant. It consists of two separate branches and is characterized by its center, vertices, foci, and asymptotes, all of which play roles in defining its shape and orientation.

Vertices

Vertices are key points on a hyperbola that lie on the transverse axis and are the nearest points to the center along that axis. The distance from the center to a vertex is critical for forming the standard equation of the hyperbola, as it helps determine the scale and orientation of its branches.

Eccentricity

Eccentricity is a measure of how much a conic section deviates from being circular. For hyperbolas, the eccentricity is greater than one. It is defined as the ratio of the distance from the center to a focus over the distance from the center to a vertex. This value is essential for understanding the shape and spread of the hyperbola.

Cartesian Equation

A Cartesian equation expresses a conic section's set of points in the coordinate plane using x and y variables. For hyperbolas, standard forms of the equation incorporate the center, transverse and conjugate axes, and are adjusted based on the orientation of the hyperbola. This equation provides a precise algebraic description of its geometry.

Transcript

00:03 In this problem, we're given the foci and the vertices for a hyperbola and asked to find the equation of the hyperbola.

00:13 The first thing we want to recognize is that the vertical coordinates for both the vertices and foci are the same values.

00:24 As a result of the vertical components being the same values, we know that we have a horizontal hyperbola.

00:34 So for horizontal hyperbola, we know the general form of the equation that we want to find.

00:47 And that form is x minus x -subs -0 quantity squared divided by a squared minus y -subs -0 quantity squared, divided by b -squared equals 1.

01:08 Next, because we have a hyperb that's in the horizontal direction, we know the foci take on the following format, x -0 minus c comma y -0, and x -0 plus c, comma, y -0.

01:27 Comparing corresponding components, we can obtain the following equations.

01:34 X of 0 minus c equals negative 2, and x -0 plus c equals 6.

01:50 We can also compare, the second component, to see that y sub 0 equals 2.

02:03 From the first components, we obtained a system of equations, two equations and two unknowns.

02:10 We can solve fairly quickly using the elimination or addition method, recognizing that the coefficients of c are already the opposite of one another.

02:19 So we can go ahead and add the two equations together to eliminate c, x of 0 plus x of 0 is 2 times x of 0.

02:27 Negative c plus c is 0 equals negative 2 plus 6 which is 4.

02:34 Divide in both sides by 2, we get x of 0 equals 2.

02:39 Next we want to back substitute into either of the previous equations to solve for c.

02:48 So we have x of 0 plus c equals 6.

02:53 Our x of 0 is 2.

02:57 Solving for c by subtracting 2 on both sides, we obtain c equals 4...