Question

The transverse axis of the hyperbola lies on the line y=-3 and has length 6, and the conjugate axis lies on the line x=2 and has length 8.

Answer 1

The general equation of a hyperbola is represented as: $\mathbf{\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1}$

The equation of the hyperbola is: $\mathbf{\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1}$

Recall that:

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

The hyperbola lies on $\mathbf{y = -3}$

This means that: $\mathbf{k =y}$

So, we have:

$\mathbf{k =-3}$

The length on the y-axis is 6.

So:

$\mathbf{a = \frac 62}$

$\mathbf{a = 3}$

Similarly, the hyperbola lies on $\mathbf{x = 2}$

This means that: $\mathbf{h =x}$

So, we have:

$\mathbf{h =2}$

The length on the x-axis is 8.

So:

$\mathbf{b = \frac 82}$

$\mathbf{b = 4}$

At this point, we have:

$\mathbf{k =-3}$

$\mathbf{a = 3}$

$\mathbf{h =2}$

$\mathbf{b = 4}$

Substitute these values in $\mathbf{\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1}$

$\mathbf{\frac{(x - 2)^2}{3^2} - \frac{(y -- 3)^2}{4^2} = 1}$

$\mathbf{\frac{(x - 2)^2}{3^2} - \frac{(y + 3)^2}{4^2} = 1}$

Evaluate the exponents

$\mathbf{\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1}$

Hence, the equation of the hyperbola is: $\mathbf{\frac{(x - 2)^2}{9} - \frac{(y + 3)^2}{16} = 1}$

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