right answer?

[jaredroy23]

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?

 

[dagr8est]

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.

 

[soroban]

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.