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anonymous
 one year ago
Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form.
anonymous
 one year ago
Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. Put the equation in standard form.

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Hero
 one year ago
Best ResponseYou've already chosen the best response.0If you are given the focus \((x_1, y_1)\) and directrix \((x_2, y_2)\) of a parabola, then you can insert those points in to the distance formula: \((x  x_1)^2 + (y  y_1)^2 = (x  x_2)^2 + (y  y_2)^2 \). Then simplify and solve for y to find the standard form of the parabola.

Hero
 one year ago
Best ResponseYou've already chosen the best response.0You are basically given two points: Focus: \((x_1, y_1) = (2, 4)\) Directrix \((x_2, y_2) = (x,6)\) Insert the points into the formula then simplify and isolate y. Try to think about what should be done first. You can ask questions if you're confused about anything.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea but what is the directrix? like what do i put for that ?

Hero
 one year ago
Best ResponseYou've already chosen the best response.0The directrix is a horizontal line y = 6. When you express that line as a point it becomes \((x, 6)\). What that means is, no matter the value of x, y will always be 6 at every point on the line.

Hero
 one year ago
Best ResponseYou've already chosen the best response.0Basically, it's just a matter of inserting the points into the formula and then simplifying. Have you tried inserting the points? Insert what you can and then post what you've done so far here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried but when it says yy2 what do i put for that ?

Hero
 one year ago
Best ResponseYou've already chosen the best response.0Please post everything you inserted for the entire formula. We'll go over anything you're stuck on afterwards. You can use the draw button to post your steps if you want.
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