document.write( "Question 719647: Can someone please help..
\n" ); document.write( "a hyperbola has vertices (+-5,0) and one focus (6,0) what is the standard form equation of the hyperbola. \n" ); document.write( "

Algebra.Com's Answer #441463 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
The standard forms for the equation of a hyperbola is:
\n" ); document.write( " $\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\"$ for horizontal hyperbolas; or
\n" ); document.write( " $\"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\"$ for vertical hyperbolas

\n" ); document.write( "The center of a hyperbola is halfway between the vertices. Halfway between (or the midpint of) (-5, 0) and (5, 0) is (0, 0). So the center of the hyperbola is (0, 0). In the standard form, these coordinates are h and k. So h = 0 and k = 0.

\n" ); document.write( "The distance between the center and each vertex is called \"a\". The distance between (0, 0) and either vertex is 5. So a = 5.

\n" ); document.write( "The distance between the center and each focus is called \"c\". The distance between (0, 0) and the one focus we know is 6. So c = 6.

\n" ); document.write( "What we need now is \"b\". The value of b is connected to the values of a and c by the equation:
\n" ); document.write( " $\"c%5E2=a%5E2%2Bb%5E2\"$
\n" ); document.write( "Using the values we have for and and c this becomes:
\n" ); document.write( " $\"%286%29%5E2+=+%285%29%5E2+%2B+b%5E2\"$
\n" ); document.write( "The standard form only requires that we know what $\"b%5E2\"$ is so we will solve for that:
\n" ); document.write( " $\"36+=+25+%2B+b%5E2\"$
\n" ); document.write( " $\"11+=+b%5E2\"$

\n" ); document.write( "And finally we use the values we have found to write the standard form equation. But which form will we use? The horizontal one or the vertical one? The answer: The horizontal one. Our hyperbola is horizontal because the vertices are on the same horizontal line. (If you can't picture (-5, 0) and (5, 0) in your head, plot them on a graph. You will find that they are to the left and right of each other.) So we will use the standard form for horizontal hyperbolas:
\n" ); document.write( " $\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1\"$
\n" ); document.write( "Into this we substitute the values we have found:
\n" ); document.write( " $\"%28x-%280%29%29%5E2%2F%285%29%5E2-%28y-%280%29%29%5E2%2F%2811%29=1\"$
\n" ); document.write( "which simplifies to:
\n" ); document.write( " $\"x%5E2%2F25-y%5E2%2F11=1\"$ \n" ); document.write( "

\n" );