William W. answered • 12/06/20

Math and science made easy - learn from a retired engineer

The length of the latus rectum is "4p" where "p" is the distance between the vertex and focus or between the vertex and the directrix. Also "1/4p" is the multiplier in front of the squared term (aka, the "a" value of the quadratic).

If the axis of the parabola is parallel to the y-axis then this is a "y^{2} = x" type quadratic. And so we can write a generic form as x = a(y - k)^{2} + h but, since the latus rectum is 6 you can say it will be:

x = 1/6(y - k)^{2} + h

Now, you can plug in the two points to get 2 equations in 2 unknowns allowing you to solve for "h" and "k":

(A) -1 = 1/6((-3) - k)^{2} + h

(B) 2 = 1/6((3/2) - k)^{2} + h

But these are both quadratic equations and require a lot of algebraic manipulation to solve so I think it would be easier to use the generic form x = ay^{2} + by + c making the two equations:

(1) -1 = 1/6(-3)^{2} + b(-3) + c or -3b + c = -5/2

(2) 2 = 1/6(3/2)^{2} + b(3/2) + c or 3b + 2c = 13/4

Using Elimination, add the equations:

-3b + c = -5/2

3b + 2c = 13/4

-------------------

3c = 3/4

c = 1/4

Plugging this back in we get b = 11/12

So the equation is x = 1/6y2 + 11/12y + 1/4

Yo can complete the square to get this into vertex form and it turns out to be x = 1/6(y + 11/4)^{2} - 97/96

L B.

Thanks a lot! :)12/12/20