Please help me with a Y-Intercept?

Question

I want to know if this is right

the Y-intercept of 4xE2 - yE2=9

At the end I got

Square root of -yE2 = Square root of 9

What I did was
get the Square root of 9 which is 3 from the right side and from the left side eliminated the exponent 2 and the square root of -y and I got -y is equal to -3


my final answer was -3


Can you tell me if this is right....

Answer 1

Sorry, but it's not. Your equation has no y-intercepts.

To find a y-intercept, substitute x=0 into your equation:

4(0)^2-y^2=9
-y^2=9

Multiply by -1:

y^2=-9

The square of a real number can never be negative, so this equation has no real solution. In practical terms, this means that the graph of 4x^2-y^2=9 does not cross the y-axis.

Answer 2

Solve for y:
-y^2 = 9 - 4x^2
y^2 = 4x^2 -9 and clearly the RHS must be >= 0, which means that 4x^2 >= 9
or x^2 >= 9/4 or x >= +-3/2
There is no y-intercept because x can never get to 0. If you plot this out, you will see that the equation is undefined in the region
-3/2 <= x <= 3/2

Answer 3

For a graph to have a y-intercept, x must be zero. But in your equation, x could never be zero, because then 0 - y squared would equal 9 and that's impossible since y squared must be >= 0. So there is no y - intercept.

Answer 4

The line intercept is when the curve crosses the y axis
=> x =0 => y*y = -9 => y = 3i or y = -3i
Unless you have a -i axis the curve does not intercept

Answer 5

If I understand this,
4x^2 - y^2 = 9
Is the y-intercept when x = 0?
- y^2 = 9
y^2 = -9
y = sqrt(-9)
y = 3i

So, I'd say no, you aren't right.