document.write( "Question 1076007: find the equation of normal to the curve 3ay2= x2(x+a) at (2a.2a \n" ); document.write( "

Algebra.Com's Answer #690696 by Fombitz(32388) $\"\"$ $\"About$

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$\"3ay%5E2=x%5E2%28x%2Ba%29\"$
\n" ); document.write( "Implicitly differentiate both sides,
\n" ); document.write( " $\"6aydy=x%5E2dx%2B%28x%2Ba%29%282xdx%29\"$
\n" ); document.write( " $\"6aydy=%28x%5E2%2B2x%28x%2Ba%29%29dx\"$
\n" ); document.write( " $\"dy%2Fdx=%28x%5E2%2B2x%28x%2Ba%29%29%2F%286ay%29\"$
\n" ); document.write( "When $\"x=2a\"$ ,
\n" ); document.write( " $\"dy%2Fdx=%28%282a%29%5E2%2B2%282a%29%282a%2Ba%29%29%2F%286a%282a%29%29\"$
\n" ); document.write( " $\"dy%2Fdx=%284a%5E2%2B12a%5E2%29%2F%2812a%5E2%29\"$
\n" ); document.write( " $\"dy%2Fdx=%2816a%5E2%29%2F%2812a%5E2%29\"$
\n" ); document.write( " $\"dy%2Fdx=4%2F3\"$
\n" ); document.write( "The normal is perpendicular to the tangent and the slopes are negative reciprocals,
\n" ); document.write( " $\"m%5BN%5D%2A%284%2F3%29=-1\"$
\n" ); document.write( " $\"m%5BN%5D=-3%2F4\"$
\n" ); document.write( "So then using the point-slope form,
\n" ); document.write( " $\"y-2a=-%283%2F4%29%28x-2a%29\"$
\n" ); document.write( " $\"4%28y-2a%29=-3%28x-2a%29\"$
\n" ); document.write( " $\"4y-8a=-3x%2B6a\"$
\n" ); document.write( " $\"highlight%283x%2B4y=14a%29\"$ \r
\n" ); document.write( "
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