Question 1026748
.
giving reasons, identify the following conic sections
i)16y=-4x^2+64
ii)4-4x-y=y^2+x^2
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<pre>
1. {{{16y}}} = {{{-4x^2+64}}}.

   This equation is equivalent to

   {{{y}}} = {{{-(1/4)*x^2 + 4)}}},   or

   {{{y}}} = {{{-(1/4)*(x^2 - 16)}}}.

   It is the equation of a parabola opened downward and symmetric about the vertical axis x = 0.

   Its vertex is V = (0, 4).  y-intercept is y = 4.  x-intercepts are x = -4 and x = 4.

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  <TR>
  <TD> 

{{{graph( 330, 330, -6.5, 6.5, -6.5, 6.5,
          -(1/4)*(x^2 - 16)
)}}}


Plot {{{y}}} = {{{-(1/4)*(x^2 - 16)}}}.

  </TD>
  </TR>
</TABLE></pre><pre>2. {{{4-4x-y}}} = {{{y^2+x^2}}}.

   This equation is equivalent to 

   {{{x^2 + 4x + y^2 + y}}} = {{{4}}}.

   Complete the squares in x and y separately. You will get an equivalent equation

   {{{(x + 2)^2 -4 + (y + (1/2))^2 - (1/4)}}} = {{{4}}},   or

   {{{(x + 2)^2 + (y + (1/2))^2}}} = {{{4 + 4 + (1/4)}}},   or

   {{{(x + 2)^2 + (y + (1/2))^2}}} = {{{33/4}}}.

   It is the equation of the circle with the center at  (x,y) = (-2,{{{-1/2}}})  and the radius of  {{{sqrt(33)/2}}}.

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  <TR>
  <TD> 

{{{graph( 330, 330, -6.5, 6.5, -6.5, 6.5,
            -1/2 + sqrt(33/4 - (x+2)^2),
            -1/2 - sqrt(33/4 - (x+2)^2))
)}}}


Plot {{{(x + 2)^2 + (y + (1/2))^2}}} = {{{33/4}}}.

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</TABLE></pre>