QuestionFind the orientation, center, vertices, conjugate axis ends, foci, and lengths of the axes for $\frac{y^{2}}{16}-x^{2}=1$ . Graph it.

Studdy Solution

STEP 1

Assumptions1. The given equation is of a hyperbola $\frac{y^{}}{16}-x^{}=1$ . The standard form of a hyperbola with vertical orientation is $\frac{(y-k)^{}}{a^{}}-\frac{(x-h)^{}}{b^{}}=1$ , where (h,k) is the center, a is the semi-transverse axis, and b is the semi-conjugate axis.

STEP 2

Identify the center of the hyperbola. The center is given by the point (h, k). In our equation, h and k are both0.
$Center = (h, k) = (0,0)$

STEP 3

Identify the semi-transverse axis (a) and the semi-conjugate axis (b). In our equation, $a^{2} =16$ and $b^{2} =1$ . Therefore, a = and b =1.
$a = \sqrt{16} =$ $b = \sqrt{1} =1$

STEP 4

Find the vertices of the hyperbola. The vertices are given by the points (h, k±a). $Vertices = (h, k \pm a) = (0,0 \pm4) = (0, -4), (0,4)$

STEP 5

Find the ends of the conjugate axis. These are given by the points (h±b, k).
$Ends\, of\, conjugate\, axis = (h \pm b, k) = (0 \pm1,0) = (-1,0), (1,0)$

STEP 6

Find the foci of the hyperbola. The foci are given by the points (h, k±c), where $c^{2} = a^{2} + b^{2}$ .First, find c.
$c = \sqrt{a^{2} + b^{2}} = \sqrt{16 +1} = \sqrt{17}$ Then, find the foci.
$oci = (h, k \pm c) = (0,0 \pm \sqrt{17}) = (0, -\sqrt{17}), (0, \sqrt{17})$

STEP 7

Find the length of the transverse axis. The transverse axis is2a.
$Length\, of\, transverse\, axis =2a =2 \times4 =$

STEP 8

Find the length of the conjugate axis. The conjugate axis is2b.
$Length\, of\, conjugate\, axis =2b =2 \times1 =2$

STEP 9

The orientation of the hyperbola is vertical because the y-term is positive in the equation of the hyperbola.
$Orientation = Vertical$

STEP 10

To graph the equation, plot the center, vertices, and foci. Draw the transverse axis through the vertices and the conjugate axis through the ends of the conjugate axis. Sketch the hyperbola opening upwards and downwards through the vertices and asymptotes through the center, with slopes of ±a/b.

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