1. Equation of hyperbola

write an equation of hyperbola having 3/2 as eccentricity. the point (1,-3) as focus and the line y=2 as its corresponding directrix

2. Equation of hyperbola: $\frac{(y-\beta)^2}{b^2}-\frac{(x-\alpha)^2}{a^2}=1$

Eccentricity: $e=\sqrt{1+\frac{b^2}{a^2}}$

Foci: $(\alpha, \beta\pm ae)$

Equation of directrices: $y-\beta=\pm \frac{a}{e}$

You have to solve $\alpha,\beta,a,b$ from this.

write an equation of hyperbola having 3/2 as eccentricity. the point (1,-3) as focus and the line y=2 as its corresponding directrix
Here comes a slightly different approach:

1. Let P denotes an arbitrary point on the hyperbola, r the distance of P to the focus F and l the distance of P to the directrix. Then

$\dfrac rl = e~\implies~r = e \cdot l$

2. Plug in the values you know:

$\sqrt{(x-1)^2+(y+3)^2} = 1.5(y+2)$

3. Simplify and re-arrange so that you'll get the equation of the hyperbola in standard form. I've got:

$-\dfrac{(x-1)^2}{\frac95}+\dfrac{\left(y+\frac65\right)^2}{\l eft(\frac56\right)^2}} = 1$

4. Originally Posted by earboth
Here comes a slightly different approach:

1. Let P denotes an arbitrary point on the hyperbola, r the distance of P to the focus F and l the distance of P to the directrix. Then

$\dfrac rl = e~\implies~r = e \cdot l$

2. Plug in the values you know:

$\sqrt{(x-1)^2+(y+3)^2} = 1.5(y+2)$

3. Simplify and re-arrange so that you'll get the equation of the hyperbola in standard form. I've got:

$-\dfrac{(x-1)^2}{\frac95}+\dfrac{\left(y+\frac65\right)^2}{\l eft(\frac56\right)^2}} = 1$
Thanks,but i get (y-6)^2/36 -(x-1)^2/45=1 when i used PF= e (PD) I wrong or not?

$\sqrt{(x-1)^2+(y+3)^2}=1.5(y-2)$