Math  /  Geometry

QuestionUse vertices and asymptotes to graph the hyperbola. Locate the foci and find the equations of th x29y2100=1\frac{x^{2}}{9}-\frac{y^{2}}{100}=1
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The foci is/are at the point(s) (109,0),(109,0)(\sqrt{109}, 0),(-\sqrt{109}, 0) (Type an ordered pair. Type an exact answer, using radicals as needed. Use a comma to separat The equation of the asymptote with the positive slope is \square The equation of the asymptote with (Simplify your answers. Use integers or fractions for any rumbers in the equation.) w an example Get more help Clear 45F45^{\circ} \mathrm{F}

Studdy Solution

STEP 1

What is this asking? We need to graph the hyperbola given by its equation, find its foci, and determine the equations of its asymptotes. Watch out! Don't mix up the formulas for horizontal and vertical hyperbolas!
Also, remember that asymptotes are lines, so they need equations, not just single numbers.

STEP 2

1. Analyze the equation
2. Find the foci
3. Find the asymptotes
4. Graph the hyperbola

STEP 3

Alright, let's **do this**!
We have the hyperbola equation x29y2100=1.\frac{x^2}{9} - \frac{y^2}{100} = 1.This tells us a few things right away.
Since the x2x^2 term is positive, this is a **horizontal hyperbola**, centered at the **origin** (0,0)(0, 0).

STEP 4

We can see that a2=9a^2 = 9, so a=3a = 3.
This means the vertices are at (3,0)(-3, 0) and (3,0)(3, 0).
Also, we have b2=100b^2 = 100, so b=10b = 10.
Remember, bb helps us find the asymptotes and the foci.

STEP 5

For a horizontal hyperbola, the foci are at (±c,0)(\pm c, 0), where c2=a2+b2c^2 = a^2 + b^2.
Let's plug in our values: c2=9+100=109.c^2 = 9 + 100 = 109. So, c=109c = \sqrt{109}.

STEP 6

Therefore, the **foci** are at (109,0)(\sqrt{109}, 0) and (109,0)(-\sqrt{109}, 0).
Awesome!

STEP 7

The equations of the asymptotes for a horizontal hyperbola centered at the origin are given by y=±baxy = \pm \frac{b}{a}x.
We already know that a=3a = 3 and b=10b = 10.

STEP 8

Plugging in these values, we get y=103xy = \frac{10}{3}x and y=103xy = -\frac{10}{3}x.
These are the equations of our **asymptotes**!

STEP 9

First, **plot the vertices** at (3,0)(-3, 0) and (3,0)(3, 0).

STEP 10

Next, **draw the asymptotes**.
They pass through the origin and have slopes of 103\frac{10}{3} and 103-\frac{10}{3}.

STEP 11

Now, **sketch the hyperbola**.
The branches of the hyperbola approach the asymptotes as xx gets larger or smaller.
Make sure the hyperbola passes through the vertices.

STEP 12

The foci are at (109,0)(\sqrt{109}, 0) and (109,0)(-\sqrt{109}, 0).
The equation of the asymptote with the positive slope is y=103xy = \frac{10}{3}x.
The equation of the asymptote with the negative slope is y=103xy = -\frac{10}{3}x.
The graph of the hyperbola can be sketched using the vertices and asymptotes.

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