equation of a parabola: 3x^2-6(x+y)=10
Find Vertex, Focus, and Directrix

Question

equation of a parabola: 3x^2-6(x+y)=10
Find Vertex, Focus, and Directrix

campbell_st · Answer

well it'd rewrite it as 

$$3x^2 -6x -6y = 10~~~ or ~~~6y = 3x^2 - 6x - 10$$

then I'd look at factoring the right hand side

$$6y = 3(x^2 - 2x) - 10$$

and complete the square in x

$$6y = 3(x^2 -2x + 1) - 10 - 1$$

which becomes 

$$6y = 3(x - 1)^2 - 11$$


I'd use the general form

$$(x - h)^2 = 4a(y - k)$$

(h, k) is the vertex and a is the focal length, a , ... dirextrix is y = k - a anf focus is (h, k + a)

to find a,$$6y + 11 = 3(x - 1)^2$$
so 

$$2y + \frac{11}{3} = (x -1)^2$$

or 

$$2(y + \frac{11}{6}) = (x - 1)^2$$


lastly you need the focal length a

you know  4a = 2 

so a = 1/2 

now you can find the focus and directrix

vegan555 · Answer

thank you so much!!