Chapter 10: Problem 21
Solve. $$\begin{aligned} &x^{2}+y^{2}=4\\\ &16 x^{2}+9 y^{2}=144 \end{aligned}$$
Short Answer
Expert verified
There are no real solutions for the given system of equations.
Step by step solution
01
- Rewrite the first equation
The first equation is given by: \[x^{2} + y^{2} = 4\] This represents a circle with radius 2.
02
- Rewrite the second equation
The second equation is given by: \[16x^{2} + 9y^{2} = 144\] Divide both sides by 144 to simplify: \[\frac{16x^{2}}{144} + \frac{9y^{2}}{144} = 1\] This simplifies to: \[\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\] Which represents an ellipse with semi-major axis 4 and semi-minor axis 3.
03
- Express one variable in terms of the other
From the first equation, express y in terms of x: \[x^{2} + y^{2} = 4\] \[y^{2} = 4 - x^{2}\] \[y = \pm\sqrt{4 - x^{2}}\]
04
- Substitute y into the second equation
Substitute y from the first equation into the second equation: \[16x^{2} + 9(4 - x^{2}) = 144\] Distribute and combine like terms: \[16x^{2} + 36 - 9x^{2} = 144\] \[7x^{2} + 36 = 144\] Subtract 36 from both sides: \[7x^{2} = 108\] Divide by 7: \[x^{2} = \frac{108}{7}\] \[x = \pm\sqrt{\frac{108}{7}}\]
05
- Simplify the solution for x
Simplify x further: \[x = \pm\sqrt{\frac{108}{7}} = \pm\frac{6\sqrt{3}}{\sqrt{7}}\] Rationalize the denominator: \[x = \pm\frac{6\sqrt{21}}{7}\]
06
- Find y using x
Using the values of x to find y: \[y^{2} = 4 - x^{2}\] For \(x = \frac{6\sqrt{21}}{7}\), \[y^{2} = 4 - \left(\frac{6\sqrt{21}}{7}\right)^{2}\] Simplify y: \[y^{2} = 4 - \frac{108}{7}\] \[y^{2} = \frac{28}{7} - \frac{108}{7}\] \[y^{2} = \frac{-80}{7}\] Since y^{2} cannot be negative, there are no real solutions for y in this case.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Circle
A circle is one of the simplest forms of a conic section. In this context, a conic section is a curve obtained by intersecting a cone with a plane. For a circle, this intersection occurs at a right angle to the axis of the cone.
The standard equation for a circle centered at the origin is given by: \[x^2 + y^2 = r^2 \] where r is the radius of the circle. In our exercise, the first equation is \[x^2 + y^2 = 4 \] which represents a circle with radius 2 (since \(\text{r}^2 = 4 \) and \(\text{r} = \text{sqrt}(4) = 2\)).
Here's a quick recap of its main properties:
The standard equation for a circle centered at the origin is given by: \[x^2 + y^2 = r^2 \] where r is the radius of the circle. In our exercise, the first equation is \[x^2 + y^2 = 4 \] which represents a circle with radius 2 (since \(\text{r}^2 = 4 \) and \(\text{r} = \text{sqrt}(4) = 2\)).
Here's a quick recap of its main properties:
- The center is at (0,0).
- All points on the circle are at an equal distance (radius) from the center.
Ellipse
An ellipse is another type of conic section. It is often described as a stretched-out circle and can be formed by slicing a cone at an angle not perpendicular to its base.
The standard equation for an ellipse centered at the origin is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where a and b are the lengths of the semi-major and semi-minor axes, respectively.
In the problem, we had to rewrite the second equation: \[16x^2 + 9y^2 = 144 \] By dividing the entire equation by 144, we transformed it into: \[\frac{x^2}{9} + \frac{y^2}{16} = 1 \] From this, we can identify that:
The standard equation for an ellipse centered at the origin is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where a and b are the lengths of the semi-major and semi-minor axes, respectively.
In the problem, we had to rewrite the second equation: \[16x^2 + 9y^2 = 144 \] By dividing the entire equation by 144, we transformed it into: \[\frac{x^2}{9} + \frac{y^2}{16} = 1 \] From this, we can identify that:
- The semi-major axis (a) is 4.
- The semi-minor axis (b) is 3.
System of Equations
A system of equations is simply a set of equations with multiple variables that are solved together. The solution to the system is the point(s) where the equations intersect.
In our exercise, we have a system comprising a circle and an ellipse:
In our exercise, we have a system comprising a circle and an ellipse:
- \ x^2 + y^2 = 4 \
- \16x^2 + 9y^2 = 144 \
- First, we expressed y in terms of x using the circle's equation: \[ y = \text{±sqrt}(4 - x^2) \]
- Then, we substituted y into the ellipse's equation to reduce the number of variables and solve for x.
- Finally, using the calculated values of x, we sought corresponding values of y.