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1) The general form of a parabola is 2x^2−12x−3y+12=0

What is the standard form?

 

2) The directrix of a parabola is x = 4. Its focus is (2,6).

What is the standard form?

 

 

In need of help!

 Jan 15, 2020
 #1
avatar+20254 
+3

Standard form    y = ax^2 + bx + c

   

 2x^2−12x−3y+12=0    re-arrange so y term is on one side

3y = 2x^2-12x+12        divide through by 3

y =   2/3 x^2 - 4x + 4          done.

 Jan 15, 2020
 #2
avatar+9 
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Hi, how can that be changed to \((y-k)^2=x-h\)

dominos  Jan 15, 2020
 #3
avatar+107520 
+1

I am taking this straight from EPs answer.  I have not looked if EP's answer is correct

 

\(y = \frac{2x^2}{ 3} - 4x + 4 \\ y -4= \frac{2x^2}{ 3} - 4x \\ \frac{3}{2}(y -4)= \frac{3}{2}( \frac{2x^2}{ 3} - 4x) \\ \frac{3}{2}(y -4)= x^2 - 6x \\ \frac{3}{2}(y -4)+3^2= x^2 - 6x +3^2 \\ \frac{3}{2}(y -4)+9= x^2 - 6x +9 \\ \frac{3}{2}(y -4)+9= (x-3)^2 \\ \frac{2}{3}(\frac{3}{2}(y -4)+9)= \frac{2}{3} (x-3)^2 \\ y -4+6= \frac{2}{3} (x-3)^2 \\ y+2= \frac{2}{3} (x-3)^2 \\ \frac{3}{2}(y+2)= (x-3)^2 \\ (x-3)^2=4*\frac{3}{8}(y+2)\)

 

This is a concave up parabola with vertex = (3,-2)   and focal length a=3/8

 

I have not checked my answer.

Melody  Jan 15, 2020

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