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!
+36916 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.
+118609 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.
