right answer?

jaredroy23

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Nov 9, 2005
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At how many points do the graphs of
x^2/9 + y^2/4 = 1 and
x^2/4 - y^2/9 = 1
intersect?

correct answer 2?
 
The first equation is an horizontal ellipse with verticies at (3, 0) and (-3, 0).

The second equation is a horizontal hyperbola with verticies at (2,0) and (-2, 0).

If you picture that in your head or draw it out, you will see that there are 4 intersections.
 
Hello, jaredroy23!

At how many points do the graphs of \(\displaystyle \;\frac{x^2}{9}\,+\,\frac{y^2}{4}\:=\:1\) and \(\displaystyle \;\frac{x^2}{4}\,-\,\frac{y^2}{9}\:=\:1\) intersect?

correct answer 2? . . . . sorry, no
If you sketch the graphs, you'll find four intersections.
. . The ellipse has vertices \(\displaystyle (\pm3,0)\)
. . The hyperbola has vertices <u>inside</u> the ellipse at \(\displaystyle (\pm2,0)\)


You can also solve the system of equations:

. . . \(\displaystyle [1]\;4x^2\,+\,9y^2\;=\;36\)
. . .\(\displaystyle [2]\;9x^2\,-\,4y^2\;=\;36\)

Multiply [1] by 4: \(\displaystyle \;16x^2\,+\,36y^2\;=\;144\)
Multiply [2] by 9: \(\displaystyle \;81x^2\,-\,36y^2\;=\;342\)

Add: \(\displaystyle \;97x^2\,=\,468\;\;\Rightarrow\;\;x^2\,=\,\frac{468}{97}\;\;\Rightarrow\;\;x\,=\,\pm\sqrt{\frac{468}{97}}\)

. . Then: \(\displaystyle \;y\,=\,\pm\sqrt{\frac{180}{97}}\)

See? . . . four points.
 
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