Question

: Graph the equation below, the graph is scaled 9 high and 9 wide.

X^2 – y^2/4 = 1

y^2/9 - x^2/4 = 1

y^2/2 - x^2/4 = 1

y^2 - x^2/9 = 1

Answer 1

Answer on Question #57352 - Math - Analytic Geometry

Question

Graph the equation below, the graph is scaled 9 high and 9 wide

1. $x^{2} - \frac{y^{2}}{4} = 1$

Solution

$$\frac{x^{2}}{1^{2}} - \frac{y^{2}}{2^{2}} = 1$$ $$a = 1, b = 2$$

Vertices: $(\pm 1,0)$

Co-vertices: $(0, \pm 2)$

Asymptote: $y = \pm \frac{2}{1}x$

Foci: $(\pm c, 0) = (\pm \sqrt{5,0})$

Answer:

2. $\frac{y^2}{9} - \frac{x^2}{4} = 1$

Solution

$$\frac{y^2}{3^2} - \frac{x^2}{2^2} = 1$$ $$a = 3, b = 2$$

Vertices: $(0, \pm 3)$

Co-vertices: $(\pm 2,0)$

Asymptote: $y = \pm \frac{3}{2}x$

Foci: $(0, \pm c) = (0, \pm \sqrt{13})$

Answer:

3. $\frac{y^2}{2} - \frac{x^2}{4} = 1$

Solution

$$\frac {y ^ {2}}{(\sqrt {2}) ^ {2}} - \frac {x ^ {2}}{2 ^ {2}} = 1$$ $$a = \sqrt {2}, b = 2$$

Vertices: $(0, \pm \sqrt{2})$

Co-vertices: $(\pm 2,0)$

Asymptote: $y = \pm \frac{\sqrt{2}}{2} x = \pm \frac{1}{\sqrt{2}} x$

Foci: $(0, \pm c) = (0, \pm \sqrt{6})$

Answer:

4. $y^{2} - \frac{x^{2}}{9} = 1$

Solution

$$\frac {y ^ {2}}{1 ^ {2}} - \frac {x ^ {2}}{3 ^ {2}} = 1$$ $$a = 1, b = 3$$

Vertices: $(0, \pm 1)$

Co-vertices: $(\pm 3,0)$

Asymptote: $y = \pm \frac{1}{3} x$

Foci: $(0, \pm c) = (0, \pm \sqrt{10})$

Answer:

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