document.write( "Question 147195: Write an equation in standard form for:\r
\n" ); document.write( "\n" ); document.write( "1. An ellipse with center a the origin, one vertex at (0,5) and one co-vertex at (0,2).\r
\n" ); document.write( "\n" ); document.write( "2. A parabola with vertex at the origin and directrix y=-2.\r
\n" ); document.write( "\n" ); document.write( "3. A circle with center (0,0) passing through (-3,4).\r
\n" ); document.write( "\n" ); document.write( "4. Hyperbola with vertices (8,-4) and (8,4) and foci (8,-6) and (8,6).\r
\n" ); document.write( "\n" ); document.write( "A few basic questions: What is a directrix and what is foci?
\n" ); document.write( "The other question: How would I go about figuring out the equations? What steps should I take?
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Algebra.Com's Answer #107604 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
Write an equation in standard form for:
\n" ); document.write( "1. An ellipse with center a the origin, one vertex at (0,5) and one co-vertex at (0,2).
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document.write( "Draw those points: \r\n" );
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document.write( "It is impossible to have a standard ellipse with a vertex and a \r\n" );
document.write( "co-vertex both on the y-axis.  So you have copied one of those \r\n" );
document.write( "points backward. I'll first assume the vertex should have been \r\n" );
document.write( "(5,0) instead of (0,5). Then we have this graph:\r\n" );
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document.write( "It is now easy to sketch in an ellipse with one vertex (5,0) and a co-vertex (0,2), as you see:\r\n" );
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document.write( "This ellipse is elongated horizontally like an egg resting on a \r\n" );
document.write( "table top.  Such an ellipse has the equation:\r\n" );
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document.write( "\"%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2=1\"\r\n" );
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document.write( "where \r\n" );
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document.write( "(\"h\", \"k\") is the center,\r\n" );
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document.write( "\"a=DISTANCE_FROM_CENTER_TO_VERTEX\", and\r\n" );
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document.write( "\"b=DISTANCE_FROM_CENTER_TO_CO-VERTEX\", \r\n" );
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document.write( "So in your problem, using (h,k) = (0,0)\r\n" );
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document.write( "\"%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2=1\" becomes\r\n" );
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document.write( "\"%28x-0%29%5E2%2F5%5E2+%2B+%28y-0%29%5E2%2F2%5E2=1\"\r\n" );
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document.write( "\"x%5E2%2F25+%2B+y%5E2%2F4=1\"\r\n" );
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document.write( "Next I'll assume the co-vertex should have been (2,0) \r\n" );
document.write( "instead of (0,2). Then we have this graph instead:\r\n" );
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document.write( "As you see, it is also easy to sketch in an ellipse with one vertex\r\n" );
document.write( "(0,5) and a co-vertex (2,0):\r\n" );
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document.write( "This ellipse is elongated vertically like the character zero \"0\".  It\r\n" );
document.write( "has this equation:\r\n" );
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document.write( "\"%28x-h%29%5E2%2Fb%5E2+%2B+%28y-k%29%5E2%2Fa%5E2=1\"\r\n" );
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document.write( "[Notice that \"a%5E2\" is under the term in \"y\" and that\r\n" );
document.write( "\"b%5E2\" is under the term in \"x\", which was the other\r\n" );
document.write( "way around in the case above. It is always true, though,\r\n" );
document.write( "that \"a\" and \"a%5E2\" are always greater than \"b\"\r\n" );
document.write( "and \"b%5E2\" respectively.]\r\n" );
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document.write( "where \r\n" );
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document.write( "(\"h\", \"k\") is the center,\r\n" );
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document.write( "\"a=DISTANCE_FROM_CENTER_TO_VERTEX\", and\r\n" );
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document.write( "\"b=DISTANCE_FROM_CENTER_TO_CO-VERTEX\", \r\n" );
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document.write( "So in interpreting your problem this way, we have,\r\n" );
document.write( "again using (h,k) = (0,0),:\r\n" );
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document.write( "\"%28x-h%29%5E2%2Fb%5E2+%2B+%28y-k%29%5E2%2Fa%5E2=1\" becomes\r\n" );
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document.write( "\"%28x-0%29%5E2%2F2%5E2+%2B+%28y-0%29%5E2%2F5%5E2=1\"\r\n" );
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document.write( "\"x%5E2%2F4+%2B+y%5E2%2F25=1\"\r\n" );
document.write( " \r\n" );
document.write( "-------------------------------------\r\n" );
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document.write( "2. A parabola with vertex at the origin and directrix y=-2\r\n" );
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document.write( "Such a parabola has equation:\r\n" );
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document.write( "\"%28x-h%29%5E2+=+4p%28y-k%29\" where the vertex is (\"h\", \"k\").\r\n" );
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document.write( "Since this parabola has vertex (\"0\",\"0\"), it has this\r\n" );
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document.write( "\"x%5E2=4py\"\r\n" );
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document.write( "We mark the origin and draw the directrix, in green, which is given as a\r\n" );
document.write( "horizontal line 2 units below and parallel to the x-axis.\r\n" );
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document.write( "The vertex is 2 units ABOVE the directrix.  That's 2 units upward,\r\n" );
document.write( "so therefore p = 2.  ---(If the vertex had been BELOW the directrix, we would have used p = -2)\r\n" );
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document.write( "Since the directrix is always entirely OUTSIDE the parabola,\r\n" );
document.write( "we know that the parabola opens upward like this:\r\n" );
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document.write( "So since p=2 the equation of the parabola is\r\n" );
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document.write( "\"x%5E2=4py\"\r\n" );
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document.write( "\"x%5E2=4%282%29y\"\r\n" );
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document.write( "\"x%5E2=8y\"\r\n" );
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document.write( "If you need the focus, (or focal point), it is a\r\n" );
document.write( "point inside the parabola which is just as far from\r\n" );
document.write( "the vertex as the directrix, but on the inside of\r\n" );
document.write( "the parabola.  In other words, the focus (or focal\r\n" );
document.write( "point) is (0,2) as you see in the graph below. Since\r\n" );
document.write( "you asked about 'directrix' and 'foci' in one of the \r\n" );
document.write( "other problems, you have to understand focus (or focal point).\r\n" );
document.write( "I did not mention the focus in the ellipse problems.  Perhaps\r\n" );
document.write( "I should have.  All conics have at least one focal\r\n" );
document.write( "point and a directrix.\r\n" );
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document.write( "A circle with center (0,0) passing through (-3,4). \r\n" );
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document.write( "First we'll draw it.  We draw the center and the point(-3,4):\r\n" );
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document.write( "Next we draw the radius from (0,0) to (-3,4)\r\n" );
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document.write( "Then we get a compass and draw the circle:\r\n" );
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document.write( "The equation of a circle with center (h,k) and radius r is\r\n" );
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document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\"\r\n" );
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document.write( "We know that the center is (0,0), so this becomes\r\n" );
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document.write( "\"x%5E2%2By%5E2=r%5E2\"\r\n" );
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document.write( "We can find the radius either by using the distance formula\r\n" );
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document.write( "\"d=sqrt%28++%28x%5B2%5D-x%5B1%5D%29%5E2%2B%28y%5B2%5D-y%5B1%5D%29%5E2%29\"\r\n" );
document.write( "=\"sqrt%28%28-3-0%29%5E2%2B%284-0%29%5E2%29\"\r\n" );
document.write( "=\"sqrt%28%28-3%29%5E2%2B%284%29%5E2%29\"\r\n" );
document.write( "=\"sqrt%289%2B16%29\"\r\n" );
document.write( "=\"5\"\r\n" );
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document.write( "or an easier way would be to have plugged (x,y)=(-3.4) into the\r\n" );
document.write( "equation:\r\n" );
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document.write( "\"x%5E2%2By%5E2=r%5E2\"\r\n" );
document.write( "\"%28-3%29%5E2%2B%284%29%5E2=r%5E2\"\r\n" );
document.write( "\"9%2B16=r%5E2\"\r\n" );
document.write( "\"25=r%5E2\"\r\n" );
document.write( "\"sqrt%2825%29=r\"\r\n" );
document.write( "\"5=r\"\r\n" );
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document.write( "So the equation\r\n" );
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document.write( "\"x%5E2%2By%5E2=r%5E2\"\r\n" );
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document.write( "becomes\r\n" );
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document.write( "\"x%5E2%2By%5E2=5%5E2\"\r\n" );
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document.write( "or\r\n" );
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document.write( "\"x%5E2%2By%5E2=25\"\r\n" );
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document.write( "---------------------------------------------\r\n" );
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document.write( "Hyperbola with vertices (8,-4) and (8,4) and foci (8,-6) and (8,6).\r\n" );
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document.write( "Let's draw those points:\r\n" );
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document.write( "The center of a hyperbola is the point midway between\r\n" );
document.write( "the two vertices, which is also the point midway between \r\n" );
document.write( "the two foci.  So the center of the hyperbola is (8,0), so\r\n" );
document.write( "we plot it:\r\n" );
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document.write( "The value of \"a\" is the distance from the center to a vertex.\r\n" );
document.write( "The distance between (8,0) and the vertex (8,4) is 4 units, so \"a=4\"\r\n" );
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document.write( "The value of \"c\" is the distance from the center to a focus.\r\n" );
document.write( "The distance between (8,0) and the vertex (8,6) is 6 units, so \"c=6\"\r\n" );
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document.write( "Now there is a Pythagorean relationship hyperbolas, that is,\r\n" );
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document.write( "\"c%5E2=a%5E2%2Bb%5E2\"\r\n" );
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document.write( "Substituting:\r\n" );
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document.write( "\"6%5E2=4%5E2%2Bb%5E2\"\r\n" );
document.write( "\"36=16%2Bb%5E2\"\r\n" );
document.write( "\"20=b%5E2\"\r\n" );
document.write( "\"sqrt%2820%29=b\"\r\n" );
document.write( "\"4.47=b\" approximately.\r\n" );
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document.write( "Draw horizontal line segments through each vertex,\r\n" );
document.write( "both right and left of each vertex, each equaling\r\n" );
document.write( "to \"b=4.47\"\r\n" );
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document.write( "Now complete the defining rectangle for the hyperbola:\r\n" );
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document.write( "Draw the extended diagonals of that rectangle:\r\n" );
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document.write( "Draw in the hyperbola:\r\n" );
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document.write( "Hyperbolas which open up and down have this equation:\r\n" );
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document.write( "\"%28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1\"\r\n" );
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document.write( "where {h,k) = the vertex, a = distance from center to vertex,\r\n" );
document.write( "and b= distance from vertex to nerest corner of defining \r\n" );
document.write( "rectangle.\r\n" );
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document.write( "So (h,k) = (8,0), \"a=+4\", \"b=sqrt%2820%29\"\r\n" );
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document.write( "Therefore the equation is\r\n" );
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document.write( "\"%28y-0%29%5E2%2F4%5E2-%28x-8%29%5E2%2F%28sqrt%2820%29%29%5E2=1\"\r\n" );
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document.write( "\"y%5E2%2F16-%28x-8%29%5E2%2F20=1\"\r\n" );
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document.write( "Edwin
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