Hyperbola in Standard Form and Vertices, Co– Vertices, Foci, and Asymptotes of a Hyperbola
| \(\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\) | \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\) |
| 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:
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: \((-4, 2+8\sqrt{2}), (-4, 2-8\sqrt{2})\)
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Original price was: $109.99.$54.99Current price is: $54.99.
Original price was: $109.99.$54.99Current price is: $54.99.
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