Easiest way to verify that $4x^2+y^2=1$ is an ellipse?

Question

Normally I would just divide both sides by the number $4$ because it's not good in there, but I can't do it for

$$4x^2+y^2=1$$

I must have $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

So what's the easiest way?

Answer 1

HINT:

$$4 = \frac{1}{\frac{1}{4}}$$

Answer 2

For general case, a quadratic equation of two variables $x$ & $y$ is given as $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ The above equation will represent an ellipse iff the following conditions are satisfied $$\Delta=(abc+2fgh-af^2-bg^2-ch^2)\neq0$$ $h^2-ab<0$ & $a\neq b$

As per your question, we have $a=4$, $h=0$, $b=1$, $g=0$, $f=0$ & $c=-1$ $$\Delta=(4)(1)(-1)=-4\neq0$$ & $ a\neq b$ Hence, the equation represents an ellipse.