Question

Help with classifying conics by discriminant

Answer 1

Identify all the possible values for A in the following equation that would make the conic section a hyperbola:

Ax2 − 3xy + 4y2 − 5 = 0

Answer 2

So if its a hyperbola we use:

B^2 − 4AC > 0

Answer 3

-3^2-4(1)(4)

Answer 4

9-16

Answer 5

@Nnesha

Answer 6

-7>0

Answer 7

@perl did I do something wrong since -7 is not greater than 0?

Answer 8

In general we can write this:
given the subsequent quadratic equation:
\[\Large a{x^2} + bxy + c{y^2} + dx + ey + f = 0\]
we have:
\[\Large \begin{gathered}
\Delta = {b^2} - 4ac < 0\;\left( {ellipse} \right) \hfill \\
\Delta = {b^2} - 4ac = 0\;\left( {parabola} \right) \hfill \\
\Delta = {b^2} - 4ac > 0\;\left( {hyperbola} \right) \hfill \\
\end{gathered} \]

Answer 9

Indeed then I believe I set up the equation right. I am just confused on what I did because -7 is not greater than 0 s: This would be an ellipse if so

Answer 10

if A=1 (in your equation) then \Delta =-7 <0, so your conic is an ellipse

Answer 11

ax2 − 3xy + 4y2 − 5 = 0
a= 1
b = -3
c = 4
d = 0
e = 0
f = -5

To make a hyperbola you need b^2 -4ac > 0

(-3)^2 - 4(a)(4) > 0
9 - 16a >0

9 > 16a

a < 9/16

Answer 12

ohhh so you ahve to keep A an exponent

Answer 13

`a` as a constant

Answer 14

No exponent i mean variable T_T

Answer 15

A is a coefficient

Answer 16

A is a factor

Answer 17

right, think of it as a parameter ;)

Answer 18

different specific values of `a` give you different hyperbolas. but strictly speaking x and y are the variables

Answer 19

we are finding the set of parameters 'a' that cause the conic to be hyperbola