Question 658496
<pre>
You gave the corresponding directrix as just "y".  
You have to give an equation, not just "y".  I will
guess arbitrarily that you meant "y=2".  If you meant
another number, then the principle will be the same.

                  distance from any pt. (x,y) to focus)  
eccentricity = -------------------------------------------
                distance from that pt. (x,y) to directrix)

Distance from (x,y) to focus (1,-3) = {{{sqrt((x-1)^2+(y+3)^2)}}}

Distance from (x,y) to the point (x,2) on the directrix = 

{{{sqrt((x-x)^2+(y-2)^2)}}} = {{{sqrt((0)^2+(y-2)^2)}}} = |y-2|

Use the eccentricity formula above:

                    {{{3/2}}} = {{{(sqrt((x-1)^2+(y+3)^2))/abs(y-2)}}}

Square both sides:

                    {{{9/4}}} = {{{((x-1)^2+(y+3)^2)/(y-2)^2}}}

Cross-multiply:

            9(y - 2)² = 4[(x - 1)² + (y + 3)²]

       9(y² - 4y + 4) = 4(x - 1)² + 4(y + 3)²

       9y² - 36y + 36 = 4(x - 1)² + 4(y² + 6y + 9)

       9y² - 36y + 36 = 4(x - 1)² + 4y² +24y + 36

5y² - 60y - 4(x - 1)² = 0

Factor out 5 from the first two terms:

5(y² - 60y) - 4(x - 1)² = 0

Complete the square in the first parentheses:
-60×{{{1/2}}} = -30, the (-30)² = 900. Add 900 in the parentheses
and since the parentheses has coefficient 5 we add 5·900 or 4500 
on the right:

5(y² - 60y + 900) - 4(x - 1)² = 0 + 4500

5(y - 30)² - 4(x - 1)² = 4500

Get a 1 on the right by dividing through by 4500:

{{{(5(y-30)^2)/4500}}} - {{{(4(x-1)^2)/4500}}} = 1

{{{((y-30)^2)/900}}} - {{{((x-1)^2)/1125}}} = 1

Edwin</pre>