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For all the graphs I have plotted with non zero x², y² and xy term, I always have seen a hyperbola with a rotated axis. But under any condition can it be an ellipse? If yes, how?

Sarban Bhattacharya
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2 Answers2

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The general second degree equation is $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ For this equation to represent a hyperbola $h^2>ab$ must hold and $\Delta= abc+2fgh-af^2-bg^2-ch^2\ne 0$ must also hold.

Just for example plot the equation $$2x^2-3xy-9x+y^2+6y+9=0$$ and see the result. Is this a hyperbola$?$

And no, if the equation has non zero coefficients of $x^2,y^2,xy$ then it can neither be a parabola nor a ellipse.

The above answer assumes no rotation. But if rotation of the curve is allowed, then any second degree equation will represent a parabola if $h^2=ab$ and $\Delta\ne0$ and any general second degree equation will represent an ellipse if $h^2<ab$ and $\Delta\ne0$ no matter whether there are non zero coefficients of $x^2$ and $y^2$ or not

MathStackexchangeIsNotSoBad
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  • But this comes out as a straight line which is a degenerate case of hyperbola. Other than this, is anything possible? – Sarban Bhattacharya Apr 01 '23 at 13:20
  • @SarbanBhattacharya when I answered your question I assumed no rotation... otherwise any general second degree eqn with $\Delta\ne0$ and $h^2=ab$ will represent a parabola...that will be a roated parabola – MathStackexchangeIsNotSoBad Apr 01 '23 at 13:31
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I hate standard form and always see if I can convert them to their more useful forms:

circle: (x-h)^2 + (y-v)^2 = r^2 - (h,v) are the horizontal and vertical coordinates of the circle's center

ellipse: (x-h)^2/a^2 + (y-v)^2/b^2 = 1, where a is the semimajor axis and b is the semiminor axis

parabola: 4p(y-v) = (x-h)^2, where p is the distance between the vertex and the focus, or y = a(x-h)^2 + v

hyperbola: (x-h)^2/a^2 - (y-v)^2/b^2 = 1.

See if you can algebraically manipulate it to one of those forms or if doing so is impossible for an ellipse.

Nate
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