Hyperbola in Standard Form and Vertices, Co– Vertices, Foci, and Asymptotes of a Hyperbola

Hyperbole is determined by the center, vertices, and asymptotes.

Hyperbola in Standard Form  and Vertices, Co– Vertices, Foci, and Asymptotes of a Hyperbola

The standard forms for the equation of hyperbolas are:

\(\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\) and \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

The information of each form is written in the table below:

\(\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\) \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)
Center: \((h, k)\)
Foci: \((h, k ± c)\)
Vertices: \((h, k ± a)\)
Transverse axis: \(x=h\)
(Parallel to \(y\)-axis)
Asymptotes: \(y-k=±\frac{a}{b}(x-h)\)
Center: \((h, k)\)
Foci: \((h ± c, k)\)
Vertices: \((h ± a, k)\)
Transverse axis: \((y=k)\)
(Parallel to \(x\) -axis)
Asymptotes: \(y-k=±\frac{b}{a}(x-h)\)

Hyperbola in Standard Form and Vertices, Co– Vertices, Foci, and Asymptotes of a Hyperbola – Example 1:

Find the center and foci of \(x^2+y^2+8x-4y-44=0\)

Solution:

To rewrite in standard form, first add \(44\) to both sides: \(x^2+y^2+8x-4y=44\)

Group \(x\) -variables and \(y\) -variables together: \((x^2+8x)+(y^2-4y)=44\)

Convert \(x\) and \(y\) to square form: \((x^2+8x+16)+(y^2-4y+4)=44+16+4 → (x+4)^2+(y-2)^2=64\)

Divide by \(64\): \(\frac{(x+4)^2}{64}-\frac{(y-2)^2}{64}=1\)

Then: \((h, k)=(-4, 2), a=8, b=8,\) and center is \((-4, 2)\)

Foci: \((-4, 2+c), (-4,2-c)\)

Compute \(c: c=\sqrt{8^2+8^2}=8\sqrt{2}\) then: \((-4, 2+8\sqrt{2}), (-4, 2-8\sqrt{2})\)

What people say about "Hyperbola in Standard Form and Vertices, Co– Vertices, Foci, and Asymptotes of a Hyperbola"?

No one replied yet.

Leave a Reply

X
30% OFF

Limited time only!

Save Over 30%

Take It Now!

SAVE $5

It was $16.99 now it is $11.99

Math and Critical Thinking Challenges: For the Middle and High School Student