It's given in my text book that the tangent at the origin can found out by equating to zero the lowest degree terms in $x$ and $y$. Therefore, by manipulating-

$yx^2-16y-2x^2+8=0$, the lowest degrees terms are $-16y-2x^2$, therefore- $x=\pm \sqrt{-8y}$, which is imaginary. Thus, we have a conjugate.

But the graph of $y=\frac{2x^2-8}{x^2-16}$ doesn't prove the findings.

Am I doing something wrong?

  • $\begingroup$ The lowest degree term is $-16y$ $\endgroup$ May 27, 2020 at 19:24
  • $\begingroup$ By "tangent at the origin" I assume you mean the derivative of y at (0,0)? If so, the point (0,0) is not on the curve so there is no tangent at the origin, that may explain it. $\endgroup$
    – Ty.
    May 27, 2020 at 19:25
  • $\begingroup$ $x=\pm\sqrt{-8y}$ is only imaginary if $y > 0$. $\endgroup$
    – D Ford
    May 27, 2020 at 19:28
  • $\begingroup$ When you say "singular point", do you mean you're trying to find the singularities? If so, there are two, corresponding to when $x^2 - 16 = 0$. $\endgroup$
    – D Ford
    May 27, 2020 at 19:30
  • $\begingroup$ That curve does not pass through the origin in $xy$-rectangular coordinates. $\endgroup$
    – Allawonder
    May 27, 2020 at 21:41

1 Answer 1


HINT: The singular points are the values of $x$ for which $y$ diverges. Look at the conditions when the denominator of the expression for $y$ equates to $0$, and when $x \to \pm\infty$.


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