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A system of equations is made up of an ellipse and a hyperbola.

Part A: Create the equation of an ellipse centered at the origin, with a vertical major axis of 8 units and a minor axis of 6 units. Show your work. (3 points)

Part B: Create the equation of a hyperbola centered at the origin, with a horizontal transverse axis, vertex at (–4, 0), and asymptotes of y equals plus or minus seven fourths x period Show your work. (4 points)

Part C: Determine the domain of each conic section and use it to explain why there is no solution to the system. (3 points)

Respuesta :

The equation of the ellipse and the equation of the hyperbola can be

derived from the given information and the general form of the equations.

[tex]Part \, A: The \ equation \ of \ the \ ellipse \ is; \displaystyle \frac{x^2}{3^2} + \frac{y^2}{4^2} = 1[/tex]

[tex]Part \, B: The \ equation \ of \ the \ hyperbola \ is; \displaystyle \frac{x^2}{(-4)^2} - \frac{y^2}{(-7)^2} = 1[/tex]

Part C: The domain of the ellipse is; [-3, 3]

The domain of the hyperbola is; (-∞, -4]

  • The domains of the ellipse and the hyperbola do not intersect, therefore, the system of equation has no solution.

Reasons:

Part A: The center of the ellipse = At the origin;

The vertical major axis = 8 units

The minor axis = 6 units

The general equation for an ellipse is presented as follows;

[tex]\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1[/tex]

Where;

[tex]\displaystyle a = The \ \mathbf{ semi \ major \ axis} = \frac{8}{2} = 4[/tex]

[tex]\displaystyle b = The \ semi \ \mathbf{ minor} \ axis = \frac{6}{2} = 3[/tex]

The equation of the ellipse is therefore;

  • [tex]\displaystyle \underline{\frac{x^2}{3^2} + \frac{y^2}{4^2} = 1}[/tex]

Part B: center of the hyperbola = The origin

The transverse axis = Horizontal

The vertex of the hyperbola = (-4, 0)

[tex]\displaystyle The \ asymptote \ are; \ y = \mathbf{ \pm\frac{7}{4} \cdot x}[/tex]

The general equation of an hyperbola is presented as follows;

[tex]\displaystyle \mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2}} = 1[/tex]

[tex]\displaystyle The \ asymptote \ are; \ y = \mathbf{\pm\frac{b}{a} \cdot x}[/tex]

The vertices = (a, 0), (-a, 0)

Therefore, by comparison, we have;

a = -4

b = -7

Which gives the equation of the hyperbola as follows;

  • [tex]\displaystyle \underline{ \frac{x^2}{(-4)^2} - \frac{y^2}{(-7)^2} = 1}[/tex]

The above equation can be written as follows;

[tex]\displaystyle \frac{x^2}{(4)^2} - \frac{y^2}{(7)^2} = 1[/tex]

Part C: Given that both equations are equal, we have;

The covertices of the ellipse are; (-3, 0) and (3, 0)

The domain of the ellipse = [-3, 3]

The domain of the hyperbola = x < The negative vertex

∴ The domain of the hyperbola = x < -4 = (-∞, -4]

  • The above domain of the ellipse does not extend to the hyperbola, therefore there is no solution to the system.

Learn more about ellipse and hyperbola here:

https://brainly.com/question/17193351

https://brainly.com/question/8957529

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