Question

find the standard form of the equation of the parabola with a focus at (-7,0) and a directrix at x=7. PLEASE HELP WILL GIVE MEDALS!!!!

Answer 1

How do you think your supposed to do this?

Answer 2

tbh im not sure.

Answer 3

my options are
x=(-1/28)y^2
-28y-=x^2
y^2=-14x
y=(-1/28)x^2

Answer 4

Hint:
we can apply the definition of a parabola

Answer 5

is there an equation I can use

Answer 6

let's suppose that P=(x,y) be a point which belongs to our parabola, then we can write:
\[\sqrt {{{\left( {x - \left( { - 7} \right)} \right)}^2} + {{\left( {y - 0} \right)}^2}} = \left| {x - 7} \right|\]

Answer 7

now, we have to square both sides of that equation, and solve it for x

Answer 8

but wouldn't the absolute value part just be a positive and a negative?

Answer 9

the absolute part is always positive, since at the left side, we have a square root, which, by definition, is positive.

Answer 10

that one has the wrong focus,
and directrix

Answer 11

what does the standard form look like?

Answer 12

I think that my equation is right. Please try to square both sides, and you will get one of your options.

Answer 13

okay just give me one sec

Answer 14

okay its not working my calculator sucks

Answer 15

here are your steps:

Answer 16

\[\begin{gathered}
\sqrt {{{\left( {x - \left( { - 7} \right)} \right)}^2} + {{\left( {y - 0} \right)}^2}} = \left| {x - 7} \right| \hfill \\
{\left( {x + 7} \right)^2} + {y^2} = {\left( {x - 7} \right)^2} \hfill \\
{x^2} + 49 + 14x + {y^2} = {x^2} + 49 - 14x \hfill \\
28x + {y^2} = 0 \hfill \\
x = - \frac{{{y^2}}}{{28}} \hfill \\
\end{gathered} \]

Answer 17

so how does it get to become x=-(1/28)y^2?

Answer 18

yes!

Answer 19

we have to develop the algebraic computation. I have started using the definition of a parabola, namely:
"a parabola is that curve whose points have the same distance from the focus and from the directrix".

Answer 20

okay

Answer 21

the distance of our generic point P=(x,y) from the focus is:
\[\sqrt {{{\left( {x - \left( { - 7} \right)} \right)}^2} + {{\left( {y - 0} \right)}^2}} \]

Answer 22

whereas the distance of the generic point P=(x,y) from the directrix, is:
\[\left| {x - 7} \right|\]

Answer 23

yes and then we squared each side
right?

Answer 24

that's right!

Answer 25

so the answer is x=(-1/28)y^2

Answer 26

yes!

Answer 27

Thank You so much!

Answer 28

Thank You!