Write the equation of the hyperbola with foci at (5,0)
and (-5,0) and eccentricity 5/3
Oops, that's not the right type of hyperbola. The form we need is
$\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$
or
$\displaystyle \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$.
In our case, for the 1st question,
Since the two foci lie on a horizontal line, we use this form of the hyperbola:Write the equation of the hyperbola with foci at (5,0)
and (-5,0) and eccentricity 5/3
$\displaystyle \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$.
It's also easy to see that the center = (h , k) = (0, 0).
Remember that the distance from the center to one of the foci is c, and
remember that eccentricity is $\displaystyle e = \frac{c}{a}$.
Also, use the Pythagorean relation $\displaystyle c^2 = a^2 + b^2$ to find b. I'll let you go on from here.
The two vertices lie on a vertical line, so the form of the hyperbola to use isFind an equation for the hyperbola with vertices
at (–2, 15) and (–2, –1), and having eccentricity e = 17/8
$\displaystyle \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$.
The midpoint between the two vertices is the center (h, k). The distance between the center and one of the vertices is a. Use the eccentricity formula and Pythagorean relation above to find c and b.
01
Hi Magentarita,
The general equation of a hyperbola is:
$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
For this formulation the directrix is x = a/e and foci at (ae,0). Also $\displaystyle b^2 = a^2(e^2 - 1)$. With this information
$\displaystyle 5 = a. \frac{5}{3}$
giving a = 3
Also $\displaystyle 3 = b^2(\frac{25}{9} - 1)$
giving $\displaystyle b^2 = \frac{27}{16}$
Therefore the equation is:
$\displaystyle \frac{x^2}{9} - \frac{16.y^2}{27} = 1$
$\displaystyle 3x^2 - 16y^2 = 27$
The thing is there are various formulations of the equation for a hyperbola and with the right choice of the constants, the question almost answers itself.