Problems from Equation of the vertical hyperbolas

Development:

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

b) The apexes are in $F^{\prime }(x_0,y_0-a)$ and $F(x_0,y_0+a)$. Like $a=2$, $F^{\prime }(4,-2)$ 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 $F^{\prime }(4,-2)$ and $F(4,2)$.

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

Hide solution and development