How do I graph this on my Ti-83 calculator?

Question

Basically I'm asking how do I solve for y on this equation so that I can graph it.
(x^2/16)-(y^2/9)=1
Thank you!

Answer 1

Since the TI-83 can't do symbolic manipulation (can't do algebra), you have to solve it yourself. However, this is easy:

(x^2)/16 - (y^2)/9 = 1
- (y^2)/9 = 1 - (x^2)/16
(y^2)/9 = (x^2)/16 - 1
y^2 = 9*((x^2)/16 - 1)
y = (9*((x^2)/16 - 1))^(1/2)

So, just plug in (9*((x^2)/16 - 1))^(1/2) at the "y=" prompt, and your calculator should graph it with no trouble.

Answer 2

For what it's worth, you don't really need your calculator to sketch the graph of this equation if you recognize that the equation is of the form

x^2/a^2 - y^2/b^2 = 1

which is a standard form for a hyperbola. Sketch a box around the origin with sides at x = ±a and y = ±b. In this particular case, a = 4 and b = 3. The asymptotes of the hyperbola will be extensions of the diagonals of the box and the vertices of the hyperbola will be tangent to the middles of two opposite sides of the box.

Because the term -y^2/b^2 can never be positive, the left side of this equation can never become as as large as 1 if x is zero, so the hyperbola can never touch the y axis. Therefore, in this case, the two halves of the hyperbola open outward away from the origin in the ±x direction.

Answer 3

To solve for y, first move the x term and the negative to the opposite side to get

y^(2/9) = x^(2/16)-1

Now to get rid of the exponent, raise both sides to the reciprocal power (that is, 9/2) so you get

y = (x^(2/16)-1)^(9/2).

You can then graph that.

Answer 4

if i'm not mistaken the graph will be an ellipse. Graphing what others posted will only show you the top half of the ellipse because it doesn't take into account the fact that y could be positve or negative.

Answer 5

you can try solving for y:
y^2/9 = (x^2/16-1)
y^2 = 9(x^2/16-1)
so you have to graph two equations:
y = sqrt(9*(x^2/16-1))
y = -sqrt(9*(x^2/16-1))