Problems from Equation of the vertical hyperbolas

Consider the hyperbola $$\dfrac{y^2}{2}-\dfrac{(x-4)^2}{18}=2$$, and find:

a) The center

b) Its apexes

c) The focal distance

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Development:

a) Firstly identify in the expression $$\dfrac{(y-y_0)^2}{a^2}-\dfrac{(x-x_0)^2}{b^2}=1$$ the equation. To do so, divide the equation given by $$2$$ so that the same remains in the part of the right. $$$\dfrac{y^2}{4}-\dfrac{(x-4)^2}{36}=1$$$ Next, identify: $$$\dfrac{y^2}{2^2}-\dfrac{(x-4)^2}{6^2}=1$$$ The center is in $$C(x_0,y_0)$$, therefore $$C(4,0)$$.

b) The apexes are in $$F'(x_0,y_0-a)$$ and $$F(x_0,y_0+a)$$. Like $$a=2$$, $$F'(4,-2)$$ and $$F(4,2)$$.

c) Look for the focal distance: $$c^2=a^2+b^2=2^2+6^2=4+36=40$$. Do the root: $$$c=\sqrt{40}=2\sqrt{10}$$$

Solution:

a) $$C (4,0)$$

b) The apexes are in $$F' (4,-2)$$ and $$F (4,2)$$.

c) The focal distance is $$c=\sqrt{40}=2\sqrt{10}$$

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