document.write( "Question 749494: consider the curve y=2logx, where log is the natural logarithm. let α be the tangent to that curve which passes through the origin, let P be the point of contact of α and that curve, and let m be the straight line perpendicular to the tangent α at P. We are to find the equations of the straight lines α and m and the area S of the region bounded by the curve y=2logx, the straight line m, and the x-axis
\n" ); document.write( "let t be the x-coordinate of the poin P, then t satisfies log t=(A). Hence the equation of α is
\n" ); document.write( "\"y=%28B%29%2Ax%2Fe\"
\n" ); document.write( "the equation of m is
\n" ); document.write( "\"y=-ex%2F%28C%29+%2B+e%5E2%2F%28D%29+%2B+%28E%29\"
\n" ); document.write( "thus the area S of the region is
\n" ); document.write( "\"S=%28F%29+%2B+%28G%29%2Fe\"\r
\n" ); document.write( "\n" ); document.write( "solve for A,B,C,D,E,F and G
\n" ); document.write( "

Algebra.Com's Answer #456151 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
The function is \"y=2%2Aln%28x%29\"
\n" ); document.write( "Its graph crosses the x-axis at the point where \"y=0\"
\n" ); document.write( "\"system%28y=2%2Aln%28x%29%2Cy=0%29\" --> \"2%2Aln%28x%29=0\" --> \"ln%28x%29=0\" --> \"x=1\"
\n" ); document.write( "The x-coordinate of point P is \"x=t\".
\n" ); document.write( "The slope of the tangent at point P is the value of the derivative at that point.
\n" ); document.write( "y'=\"2%2Fx\", so the slope of the tangent at \"x=t\" is \"2%2Ft\".
\n" ); document.write( "Since the line \"alpha\" tangent at P passes through the origin, its equation must be
\n" ); document.write( "\"y=%282%2Ft%29x\"
\n" ); document.write( "At point P, with \"x=t\", \"y=%282%2Ft%29t\" --> \"y=2\"
\n" ); document.write( "Since point P is on the graph of \"y=2%2Aln%28x%29\", its y-coordinate is \"y=2%2Aln%28t%29\"
\n" ); document.write( "So \"system%28y=2%2Cy=2%2Aln%28t%29%29\" --> \"2=2%2Aln%28t%29\" --> \"ln%28t%29=1\" --> \"system%28highlight%28A=1%29%2Ct=e%29\" and P is (e,2).
\n" ); document.write( "
\n" ); document.write( "Now we can find the equation of \"alpha\":
\n" ); document.write( "\"system%28y=%282%2Ft%29x%2Ct=e%29\" --> \"y=%282%2Fe%29x\" --> \"highlight%28B=2%29%0D%0A+%0D%0ASince+the+slope+of+line+%7B%7B%7Balpha\" is \"2%2Fe\",
\n" ); document.write( "line \"m\" perpendicular to \"alpha\" must have a slope of \"-1%2F%282%2Fe%29=-e%2F2\".
\n" ); document.write( "As \"m\" passes through P(e,2) its equation is
\n" ); document.write( "\"y-2=%28-e%2F2%29%28x-e%29\" --> \"y-2=%28-e%2F2%29x%2Be%5E2%2F2\" --> \"y=%28-e%2F2%29x%2Be%5E2%2F2%2B2\" --> \"y=-e%2Ax%2F2%2Be%5E2%2F2%2B2\"
\n" ); document.write( "So \"highlight%28C=2%29\", \"highlight%28D=2%29\", and \"highlight%28E=2%29\"
\n" ); document.write( "The line \"m\" crosses the x-axis at the point where \"y=0\"
\n" ); document.write( "\"system%28y=-e%2Ax%2F2%2Be%5E2%2F2%2B2%2Cy=0%29\" --> \"-e%2Ax%2F2%2Be%5E2%2F2%2B2=0\" --> \"-e%2Ax%2Be%5E2%2B4=0\" --> \"e%5E2%2B4=ex\" --> \"x=%28e%5E2%2B4%29%2Fe\" --> \"x=e%2B4%2Fe\"
\n" ); document.write( "
\n" ); document.write( "The area \"S\" of the region bounded by the curve \"y=2%2Aln%28x%29\", the straight line \"m\", and the x-axis is shown below.
\n" ); document.write( "\"graph%28300%2C300%2C-1%2C6%2C-1%2C6%2C2%2Aln%28x%29%2C-e%2Ax%2F2%2Be%5E2%2F2%2B2%29\"
\n" ); document.write( "
\n" ); document.write( "\"S\" can be can be calculated as the sum of:
\n" ); document.write( "the area below \"y=2%2Aln%28x%29\", and above the x-axis, between \"x=1\" and \"x=e\", \"int%282%2Aln%28x%29%2C+dx%2C+1%2C+e+%29=2%2Aint%28ln%28x%29%2C+dx%2C+1%2C+e+%29\"
\n" ); document.write( "plus the area below \"m\" between \"x=e\" and \"x=e%2B4%2Fe\", \"int%28%28-e%2Ax%2F2%2Be%5E2%2F2%2B2%29%2C+dx%2C+e%2C+e%2B4%2Fe+%29\"
\n" ); document.write( "
\n" ); document.write( "\"int%28%28-e%2Ax%2F2%2Be%5E2%2F2%2B2%29%2C+dx%2C+e%2C+e%2B4%2Fe+%29\" is easier than it seems.
\n" ); document.write( "It's just the area of the triangle with vertices (e,0), P(e,2), and (e+e/4,0)
\n" ); document.write( "Its base is \"4%2Fe\"; its height is \"2\", and its area is \"%281%2F2%29%2A2%2A%284%2Fe%29=4%2Fe\".
\n" ); document.write( "
\n" ); document.write( "Since \"int%28ln%28x%29%2C+dx%29=2%28x%2Aln%28x%29-x%29\",
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "So \"S=2%2B4%2Fe\" and \"highlight%28G=4%29\"
\n" ); document.write( "
\n" );