anonymous
  • anonymous
What is the equation of the quadratic graph with a focus of (2, 0) and a directrix of y = −12?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jdoe0001
  • jdoe0001
|dw:1437691840404:dw| any ideas where those two fellas are in the graph? can you graph them there?
anonymous
  • anonymous
|dw:1437691171977:dw|
anonymous
  • anonymous
right?

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jdoe0001
  • jdoe0001
yeap we also know that the vertex is equidistant from both meaning, the focus point is away from the vertex, the same amount the directrix is away from the vertex let's call that distance "p" so, that means the vertex is "half-way" between the focus and directrix the parabola opens towards the focus, the focus is above the directrix, in this case the parabola opens upwards then |dw:1437692110933:dw|
jdoe0001
  • jdoe0001
the distance from the focus to the vertex is 7, the distance from the directrix to the vertex is 7 the parabola is opening upwards, so the distance "p" is positive the squared variable is the "x" then now that we know where the vertex coordinates are at, and the "p" distance then \(\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) plug in your values, and solve for "y"
anonymous
  • anonymous
wouldn't the focus be on the x-axis, not the y-axis? since it is (2,0) not (0,2)
jdoe0001
  • jdoe0001
hmmmm hold the mayo.. .maybe .... I did mess up there, lemme recheck that
jdoe0001
  • jdoe0001
|dw:1437692548026:dw|
jdoe0001
  • jdoe0001
anyhow, the vertex is half-way, so that makes it a 0,-6 and yes, that makes the "p" distance to 6 only \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)
jdoe0001
  • jdoe0001
hmmm wait a sec.. dohh. the x coordinate can't be 0 anyhow hehe one sec
jdoe0001
  • jdoe0001
|dw:1437692768286:dw| \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 2}},{\color{blue}{ -6}})\\ {\color{purple}{ 6}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)
anonymous
  • anonymous
ok, we have the vertex, now what do we do to get the equation for it
jdoe0001
  • jdoe0001
just plug in the values in the parabola equation focus form
anonymous
  • anonymous
so it ends up as (x-2)^2 = 4*6(y-(-6))
jdoe0001
  • jdoe0001
yeap or \(\bf (x-{\color{brown}{ 2}})^2=4{\color{purple}{ (6)}}(y-{\color{blue}{ (-6)}})\implies (x-2)^2=24(y+6) \\ \quad \\ (x-2)^2=24y+144\implies \cfrac{(x-2)^2-144}{24}=y\)
anonymous
  • anonymous
Is there another way to write that out, because none of the above are answer choices
jdoe0001
  • jdoe0001
I guess one could distribute the denominator, and end up with \(\cfrac{(x-2)^2-144}{24}=y\implies \cfrac{(x-2)^2}{24}-\cfrac{\cancel{144}}{\cancel{24}}=y\implies \cfrac{(x-2)^2}{24}-6=y\)
jdoe0001
  • jdoe0001
you could also, expand the squared binomial, and simplify it with the 144, and then distribute the denominator
anonymous
  • anonymous
my anwer choices are x^2-4-35; -x^2+4+35; -1/24x^2+1/6x-35/6; and 1/24x^2-1/6-35/6
jdoe0001
  • jdoe0001
ok one sec
jdoe0001
  • jdoe0001
\(\bf \cfrac{(x-2)^2-144}{24}=y\implies \cfrac{[x^2-4x+4]-144}{24}=y \\ \quad \\ \cfrac{x^2-4x+4-144}{24}=y\implies \cfrac{x^2-4x-140}{24}=y \\ \quad \\ \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y\implies ?\)
anonymous
  • anonymous
what?
jdoe0001
  • jdoe0001
any... thing.. that confuses there?
anonymous
  • anonymous
what happens to the second and third fractions
jdoe0001
  • jdoe0001
yeah... whatever happens to those? so... simplify them, what do you get for the simplification? or 4/24 and 140/24?
anonymous
  • anonymous
oh, ok. The would simplify to 2/12 and 35/8
anonymous
  • anonymous
sorry 1/6
jdoe0001
  • jdoe0001
well... 2/12 = 1/6 and.... 140/24 simplifiefs to 35/6 thus \(\bf \cfrac{x^2}{24}-\cfrac{4x}{24}-\cfrac{140}{24}=y\implies \cfrac{1}{24}x^2-\cfrac{\cancel{4}}{\cancel{24}}x-\cfrac{\cancel{140}}{\cancel{24}}=y \\ \quad \\ \cfrac{1}{24}x^2-\cfrac{1}{6}x-\cfrac{35}{6}=y\)
anonymous
  • anonymous
oh, ok. Thanks for the help
jdoe0001
  • jdoe0001
yw

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