Ellipse Help! For what real number value of m does mx^2 + 4y^2 - 8y - 1 = 0 describes a)ellipse and b) circle?

Question

i would appreciate it if you would tell me how u got the number in a simple way as possible. thanks!
also, this is from a pre-calc honors class

Answer 1

for m= 4 circel
x^2+y^2-2y-1/4= 0 Center (0,1) r = sqrt(5/4)
for m>0 is an ellipse
Let´s see if real
Center x=(0,1)
Transport of coordinates to (0,1)
x=X
y=Y+1 so
mX^2+4(Y+1) ^2 -8(Y+1) -1 =0
mX^2+4Y^2+4-9=0
mX^2 +4Y^2 =5
so in normal form X^2/(5/m)+Y^2/(5/4)=1
so all m>0 we have an ellipse.m= 4 a circle

Answer 2

mx^2 + 4y^2 - 8y - 1 = 0
For the above to be a circle the coefficient of y^2 and x^2 must be equal, m =4
For the above to be an ellipse the coefficients must have the same sign, but not be equal.
With that said, you still should be able to complete the square to put your equation in the form.
A(x - h)^2 + B(y - k)^2 = C and have A, B and C all have the same sign. For the equation to be a circle or ellipse. (C≠0)
Later you will study hyperbolas and cases where there are two intersecting lines or the graph will be only a point