Question

the hyperbolas \\(\frac{x^2}{9} - \frac{y^2}{4} = 1\\) and \\(\frac{y^2}{36} - \frac{x^2}{25} = 1\\) are graphed in the standard \\((x,y)\\) coordinate plane. which of the following equations is an ellipse that intersects all 4 vertices of the hyperbolas?
graph of hyperbolas with y and x axes, origin o
f. \\(\frac{x^2}{9} + \frac{y^2}{36} = 1\\)
g. \\(\frac{x^2}{25} + \frac{y^2}{4} = 1\\)
h. \\(\frac{x^2}{25} + \frac{y^2}{9} = 1\\)
j. \\((x - 9)^2 + (y - 36)^2 = 1\\)
options: f, g, h, j

Explanation:

Step1: Find vertices of hyperbolas

For hyperbola \(\frac{x^2}{9}-\frac{y^2}{4}=1\) (horizontal transverse axis), vertices are \((\pm 3, 0)\) (since \(a^2 = 9\Rightarrow a = 3\)).
For hyperbola \(\frac{y^2}{36}-\frac{x^2}{25}=1\) (vertical transverse axis), vertices are \((0, \pm 6)\) (since \(a^2 = 36\Rightarrow a = 6\)).

Step2: Check ellipse equations

Ellipse standard form: \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (vertical major axis) or \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) (horizontal major axis), where \(a > b\).

• Option F: \(\frac{x^2}{9}+\frac{y^2}{36}=1\). Vertices of ellipse: \((0, \pm 6)\) (vertical) and \((\pm 3, 0)\) (horizontal). Check if hyperbola vertices lie on it:
• For \((\pm 3, 0)\): \(\frac{(\pm 3)^2}{9}+\frac{0^2}{36}=\frac{9}{9}=1\), satisfies.
• For \((0, \pm 6)\): \(\frac{0^2}{9}+\frac{(\pm 6)^2}{36}=\frac{36}{36}=1\), satisfies.
• Option G: \(\frac{x^2}{25}+\frac{y^2}{4}=1\). Check \((0, \pm 6)\): \(\frac{0}{25}+\frac{36}{4}=9

eq1\), fails.

• Option H: \(\frac{x^2}{25}+\frac{y^2}{9}=1\). Check \((0, \pm 6)\): \(\frac{0}{25}+\frac{36}{9}=4

eq1\), fails.

• Option J: Circle \((x - 9)^2+(y - 36)^2=1\), not ellipse, and vertices \((\pm 3, 0)\), \((0, \pm 6)\) don’t lie on it.

Answer:

F. \(\frac{x^2}{9}+\frac{y^2}{36}=1\)