The document provides step-by-step work to solve problems related to a hyperbola given by the equation 16y^2 - 9x^2 + 32y - 90x - 353 = 0. It shows how to write the hyperbola in standard form, graph it, determine the domain and range, identify the conjugate and transverse axes, and write the equation of the asymptotes.

1. Sergio
Wendy
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2. Unit
The Hyperbola
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3. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
b) Graph the hyperbola.
c) State the domain, and range.
d) State the conjugate axis.
e) Give the asymptote equation.
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4. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
= Standard form
4

5. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
Frist off, we rewrite the equation in the form of
where all the y's and x's are together.
don't forget the
don't forget the
brackets!!!
brackets!!!
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6. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
inside the brackets where i had placed the y's together and the x's
together, we then factor out so we leave the variable's square
alone.
factoring out the 16 factoring out the 9
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7. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
we now then, complete the squares. quot;complete the
squaresquot;, you have done many of those in grade 11.
multiply
multiply
remember to add the
opposite signs.
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8. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
you then bring the numbers to the other side of the equal sign.
once that's done, the total on the right side of the equal sign must
equal to 0. since the number is 144, to make it equal 0, we must divide
by... 144.
after that's done, start reducing!
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9. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
a) Give an equation in standard form.
And there you have it, in standard form.
= Standard form
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10. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
___ ­ ___
(y + 1)² (x + 5)²
=0
b) Graph the hyperbola.
9 16
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11. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
b) Graph the hyperbola.
___ ­ ___
(y + 1)² (x + 5)²
=0
9 16
okay, first off, you know the standard form of the hyperbola,
which means that you are ready to graph it.
a² = 9 ­> a = + 3 (y)
­
where do you start?
b² = 16 ­> b = + 4 (x)
­
you graph the orginal hyperbola, also, square root the
denominators of 9 and 16 which will become 3 and 4.
in that case, the 3 belongs to the y value which means you'll be
makeing the length of the hyperbola 3 units more. same thing
goes with the x value. 4 units more to the width of the hyperbola.
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12. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
___ ­ ___
(y + 1)² (x + 5)²
=0
b) Graph the hyperbola.
9 16
read as opposite signs.
x values is either shifting
left or right,
from the original graph of the hyperbola, all you have left to
y values is either shifting
do is shift left 5 and down 1.
up or down.
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13. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
c) State the domain, and range.
Domain; XER
Range; (­ , ­4) (2, )
oo oo
U
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14. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
d) State the conjugate axis, and the transverse axis.
Transverse axis = 2a
Conjugate axis = 2b
the transverse axis = 2a = 2(3) = 6
the conjugate axis = 2b = 2(4) = 8
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15. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
e) Give the asymptote equation.
Asymptote equation ­> (y ­ k ) = + ___ (x ­ h)
rise
­run
3
_
+
= (y + 1) = ­ (x + 5)
4
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16. 4. Given in general form for an hyperbola; 16y² ­ 9x² + 32y ­ 90x ­ 353 = 0.
e) Give the asymptote equation.
Asymptote equation ­> (y ­ k ) = + ___ (x ­ h)
rise
­run
3
_
+
= (y + 1) = ­ (x + 5)
4
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17. Attachments
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