# To find the singular point of $y=\frac{2x^2-8}{x^2-16}$

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?

• The lowest degree term is $-16y$ May 27, 2020 at 19:24
• 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.
– Ty.
May 27, 2020 at 19:25
• $x=\pm\sqrt{-8y}$ is only imaginary if $y > 0$. May 27, 2020 at 19:28
• When you say "singular point", do you mean you're trying to find the singularities? If so, there are two, corresponding to when $x^2 - 16 = 0$. May 27, 2020 at 19:30
• That curve does not pass through the origin in $xy$-rectangular coordinates. May 27, 2020 at 21:41

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 $$x \to \pm\infty$$.