# Hyperbolas- Please check my answers

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• Nov 23rd 2007, 04:09 PM
soly_sol
Hyperbolas- Please check my answers
Find the vertices and locate the foci for the hyperbola whose equation is given.

(x^2)/(81) - (y^2)/(4)=1

I got:
vertices: (-9,0), (9,0)
foci: (- squareroot 85, 0), (squareroot 85,0)

(y^2)/(100) - (x^2)/(16)=1

I got:
vertices: (0,-10), (0,10)
foci: (0, -2 squareroot 29), (0, 2 squareroot 29)

Match the equation to the graph.

(x^2)/(16) - (y^2)/(9)=1

I got:
http://i48.photobucket.com/albums/f2...l/f1q3g4-1.jpg

(y^2)/(16) - (x^2)/(4)=1

I got:
http://i48.photobucket.com/albums/f2...sol/f1q4g4.jpg

Find the standard form of the equation of the hyperbola satisfying the given conditions.

Foci: (-10, 0), (10, 0); vertices: (-5, 0), (5, 0)
I got: (x^2)/(25) - (y^2)/(75)=1

Foci: (0, -9), (0, 9); vertices: (0, -3), (0, 3)
I got: (y^2)/(9) - (x^2)/(72)=1

Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes.

(x^2)/(9) - (y^2)/(4)=1
I got:
http://i48.photobucket.com/albums/f2...ol/f1q7g11.jpg

(y^2)/(9) - (x^2)/(36)=1
I got:
http://i48.photobucket.com/albums/f2...sol/f1q8g5.jpg

Find the location of the center, vertices, and foci for the hyperbola described by the equation.

((x-4)^2)/(4) - ((y+4)^2)/(9)=1

I got: Center: (4, -4); Vertices: (2, -4) and (6, -4); Foci:4 -squareroot 13,-4) and (4 +squareroot 13, -4)

((y+3)^2)/(9) - ((x+4)^2)/(36)=1

I got: Center: (-4, -3); Vertices: (-4, -6) and (-4, 0); Foci: (-4, -3 - 3 squareroot 5) and (-4, -3 + 3 squareroot 5)
• Nov 23rd 2007, 10:21 PM
TKHunny
I got tired of reviewing after the first couple of pages. They look fine. Why do you doubt? Go forth with confidence.
• Nov 24th 2007, 11:07 AM
soly_sol
I doubt because I am taking math online and I have to pass. :)