document.write( "Question 1018492: Find the vertex,focus and end points of the latus rectum of the parabola y^2-6x+3=0 \n" ); document.write( "

Algebra.Com's Answer #634580 by josgarithmetic(39620) $\"\"$ $\"About$

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Knowing something about how an equation for parabola is derived is helpful.\r
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\n" ); document.write( "\n" ); document.write( "An easy transformation of the equation is $\"6%28x-1%2F2%29=y%5E2\"$ . Your parabola has a vertex at the left and parabola opens toward the right. The vertex is (1/2,0).\r
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\n" ); document.write( "\n" ); document.write( "Comparing to a more generally derived equation, you could have $\"4p%28x-1%2F2%29=y%5E2\"$ for p being distance from vertex to focus, so here, you have $\"p=3%2F2\"$ . You will expect the focus to be to the right of the vertex, for your example, and so the focus point is (2,0).\r
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\n" ); document.write( "\n" ); document.write( "(Latus Rectum is related to where the y values are in relation to the focus.)
\n" ); document.write( "(What are the y values for x=2 ?) \n" ); document.write( "

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