Question 129870
{{{12 x^2 – 3y^2 – 18x - 3y – 10 = 0}}}Start with the given equation



{{{12x^2-18x-3y^2-3y-10=0}}} Rearrange the terms



{{{12x^2-18x-3y^2-3y=+10}}} Add {{{10}}} to both sides



{{{12(x-3/4)^2-27/4-3y^2-3y=+10}}} Complete the square for the x terms



{{{12(x-3/4)^2-27/4-3(y+1/2)^2+3/4=+10}}} Complete the square for the y terms



{{{12(x-3/4)^2-3(y+1/2)^2-24/4=+10}}} Combine like terms



{{{12(x-3/4)^2-3(y+1/2)^2-6=+10}}} Reduce



{{{12(x-3/4)^2-3(y+1/2)^2=+10+6}}} Add {{{6}}} to both sides



{{{12(x-3/4)^2-3(y+1/2)^2=16}}} Combine like terms



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Notice how the equation is now in the form {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}. This means that this conic section is a hyperbola which opens up in the x direction.



So the conic section looks like this:




{{{ graph( 500, 500, -10, 10, -10, 10, sqrt((64/4-12(x-3/4)^2)/(-3))-1/2,-sqrt((64/4-12(x-3/4)^2)/(-3))-1/2) }}} Graph of  {{{12(x-3/4)^2-3(y+1/2)^2=16}}}