document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1121318>Question 1121318</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The parabola has equation y^2 = 4ax , where a is a positive constant.\r<BR>\n" );
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document.write( "The point P(at^2, 2at) lies on C.\r<BR>\n" );
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document.write( "The point S is the focus of the parabola C. The point B lies on the positive x-axis and OB = 5OS, where O is the origin.\r<BR>\n" );
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document.write( "A circle has centre B and touches the parabola C at two distinct points Q and R. Given that t cannot be zero.\r<BR>\n" );
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document.write( "Find the coordinates of the points Q and R. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #737332 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=737332\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "With the equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2=4ax ALIGN=MIDDLE  ALT=\"y%5E2=4ax\"   DISABLED_style= \"max-width:90%; height:auto;\" > with a positive, the vertex is at the origin, the parabola opens to the right, and the focus S is (a,0).  So point B is (5a,0).<br><BR>\n" );
document.write( "The circle centered at B touches the parabola in two points, Q and R.  The implication is that the circle is tangent to the parabola at those two points; it does not \"pass through\" the parabola, thus intersecting the parabola is 4 points.<br><BR>\n" );
document.write( "A generic point on the parabola has coordinates (at^2,2at) where t is a parameter.<br><BR>\n" );
document.write( "Let Q(at^2,2at) be the point in the first quadrant where the circle and parabola are tangent.  By symmetry, point R will be (at^2,-2at).<br><BR>\n" );
document.write( "To have the circle just touch the parabola at Q, we need to have radius BQ of the circle perpendicular to the tangent to the parabola at Q.<br><BR>\n" );
document.write( "The slope of BQ is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%280-2at%29%2F%285a-at%5E2%29+=+%28-2at%29%2F%285a-at%5E2%29+=+%28-2t%29%2F%285-t%5E2%29 ALIGN=MIDDLE  ALT=\"%280-2at%29%2F%285a-at%5E2%29+=+%28-2at%29%2F%285a-at%5E2%29+=+%28-2t%29%2F%285-t%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<br><BR>\n" );
document.write( "The slope of the tangent to the parabola at Q is found by evaluating the derivative at Q.<br><BR>\n" );
document.write( "Implicit differentiation gives us<br><BR>\n" );
document.write( "2yy' = 4a<BR>\n" );
document.write( "y' = (4a)/(2y) = 2a/y<br><BR>\n" );
document.write( "So the slope of the tangent to the parabola at Q is (2a)/(2at) = 1/t.<br><BR>\n" );
document.write( "We need to have the product of the slopes equal to -1:<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28-2t%29%2F%285-t%5E2%29%29%2A%281%2Ft%29+=+-1 ALIGN=MIDDLE  ALT=\"%28%28-2t%29%2F%285-t%5E2%29%29%2A%281%2Ft%29+=+-1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-2%29%2F%285-t%5E2%29+=+-1 ALIGN=MIDDLE  ALT=\"%28-2%29%2F%285-t%5E2%29+=+-1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5-t%5E2+=+2 ALIGN=MIDDLE  ALT=\"5-t%5E2+=+2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=t%5E2+=+3 ALIGN=MIDDLE  ALT=\"t%5E2+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=t+=+sqrt%283%29 ALIGN=MIDDLE  ALT=\"t+=+sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The coordinates of point Q are (at^2,2at) = (3a,2a*sqrt(3)).<br><BR>\n" );
document.write( "So the coordinates of point R are (at^2,-2at) = (3a,-2a*sqrt(3)).<br><BR>\n" );
document.write( "For the case a=1, B is (5,0) and Q is (3,2*sqrt(3)); the Pythagorean Theorem tells us the radius of the circle is 4.<BR>\n" );
document.write( "Below is a graph of the positive branches of the two curves for the case a=1.  The two graphs are <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=2%2Asqrt%28x%29 ALIGN=MIDDLE  ALT=\"y=2%2Asqrt%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-5%29%5E2%2By%5E2+=+16 ALIGN=MIDDLE  ALT=\"%28x-5%29%5E2%2By%5E2+=+16\"   DISABLED_style= \"max-width:90%; height:auto;\" > which graphs as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+sqrt%2816-%28x-5%29%5E2%29 ALIGN=MIDDLE  ALT=\"y+=+sqrt%2816-%28x-5%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The intersection of the two graphs is the point Q.<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28400%2C400%2C-2%2C8%2C-2%2C8%2C2%2Asqrt%28x%29%2Csqrt%2816-%28x-5%29%5E2%29%29 ALIGN=MIDDLE  ALT=\"graph%28400%2C400%2C-2%2C8%2C-2%2C8%2C2%2Asqrt%28x%29%2Csqrt%2816-%28x-5%29%5E2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " </SPAN>\n" );
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