Ivan Shishkin, Rye (1878)

Problems/GeometryExerciseReviewed

Equations of an ellipse ?

by Ancient Tree·
26
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Which of the following equations define an ellipse ?

  1. x2+2y2=1x^{2}+2y^{2}=-1
  2. x2+y2=2x^2+y^2=2
  3. 4x2+9y2=364x^{2}+9y^{2}=36
  4. 5x2+2y2=05x^{2}+2y^{2}=0
  5. x2y2=1x^{2}-y^{2}=1
  6. x2+2x+y2=1x^{2}+2x+y^{2}=1
  7. Show that any circle is, in particular, an ellipse.
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Solution by Ancient Tree

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  1. This equation does not admit any solution, because x2+2y2x^{2}+2y^{2} is always non-negative. So it’s not an ellipse.

  2. This is the equation of a circle of radius 2\sqrt{2}, which is a particular case of an ellipse (cf below).

  3. This is the equation of an ellipse of center (0,0)(0,0) with a=3a=3 and b=2b=2.

  4. This equation implies x=0x=0 and y=0y=0 because squares are always non-negative. So the set it defines is just a point. By convention, this is usually not considered an ellipse.

  5. This is the equation of an hyperbola, which is not an ellipse (because it is not bounded).

  6. Using x2+2x=(x+1)21x^2+2x=(x+1)^2-1, one gets (x+1)2+2y2=2(x+1)^2+2y^2=2 or also (x(1)2)2+(y01)2=1\left(\frac{x-(-1)}{\sqrt{2}}\right)^2+\left(\frac{y-0}{1}\right)^2=1 which is an ellipse.

  7. The equation of a circle is (xx0)2+(yy0)2=R2(x-x_0)^2+(y-y_0)^2=R^2 then, dividing by R2R^2, one gets the cartesian equation of an ellipse with a=ba=b.

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