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The center of a hyperbola is (-3,2). The length of the conjugate axis is 12 units, and the length of the transverse axis is 8 units. The
transverse axis is parallel to the y-axis.
What is the equation of the hyperbola in standard form?

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MathPhys

Answer:

(y − 2)² / 16 − (x + 3)² / 36 = 1

Step-by-step explanation:

The conjugate axis is the axis of symmetry.  The transverse axis is the line connecting the vertices of the hyperbola.  Since the transverse axis is parallel to the y-axis, this is a vertical hyperbola:

(y − k)² / a² − (x − h) / b² = 1

where (h, k) is the center of the hyperbola, a is half the length of the transverse axis, and b is half the length of the conjugate axis.

Here, the center is (-3, 2), a = 8/2 = 4, and b = 12/2 = 6.

(y − 2)² / 16 − (x + 3)² / 36 = 1

abidemiokin

The required equation of the parabola in standard form is expressed as [tex]\frac{(y-2)^2}{16}- \frac{(x+3)^2}{36} = 1[/tex]

The formula for finding the equation of a parabola is expressed according to the equation;

[tex]\frac{(y-k)^2}{a^2}- \frac{(x-h)^2}{b^2} = 1[/tex]

where;

(h, k) is the vertex

a and b are half the length of the transverse axis and half the length of the conjugate axis respectively

Given the following parameters

(h, k) = (-3, 2)

a = 8/2 = 4 units

b = 12/2 = 6 units

Substitute the parameters into the formula to have:

[tex]\frac{(y-2)^2}{4^2}- \frac{(x-(-3))^2}{6^2} = 1\\\frac{(y-2)^2}{16}- \frac{(x+3)^2}{36} = 1[/tex]

Hence the required equation of the parabola in standard form is expressed as [tex]\frac{(y-2)^2}{16}- \frac{(x+3)^2}{36} = 1[/tex]

Learn more here:

https://brainly.com/question/24208497

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