chrisdbest
  • chrisdbest
Help!!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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chrisdbest
  • chrisdbest
Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).
chrisdbest
  • chrisdbest
@pooja195
chrisdbest
  • chrisdbest
@Nnesha @Preetha

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More answers

jdoe0001
  • jdoe0001
so.. the vertex is a the origin the focus point is a 0. -7 care to draw those two in a cartesian plane?|dw:1437175007773:dw|
chrisdbest
  • chrisdbest
Vertex is at origin, and focus point is at 0, -7 like you said
jdoe0001
  • jdoe0001
ok... so what are the coordinates for the origin?
chrisdbest
  • chrisdbest
Isn't it 0,0?
jdoe0001
  • jdoe0001
yeap so |dw:1437175183982:dw| so.... if the parabola makes a U-turn at the 0,0 and the focus is below the vertex that means the parabola is |dw:1437175290874:dw|
jdoe0001
  • jdoe0001
so is a parabola that opens vertically meaning the "x" term is the one that's squared so \(\bf \begin{array}{llll} (y-{\color{blue}{ k}})^2=4{\color{purple}{ p}}(x-{\color{brown}{ h}}) \\ (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ h}},{\color{blue}{ k}})\\ {\color{purple}{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\ \quad \\\\ \quad \\ \textit{so we'd need to use this form }(x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\)
jdoe0001
  • jdoe0001
now... what is the distance "p", that is, from the vertex to the focus point?
chrisdbest
  • chrisdbest
hmmm
jdoe0001
  • jdoe0001
|dw:1437175751363:dw|
chrisdbest
  • chrisdbest
Wouldn't that be 7?
jdoe0001
  • jdoe0001
yeap so p is 7 the parabola is going downards, that means "p" is negative so "p" is really negative 7 or -7 so now you know what "p" is, and where the vertex is at, that is what "h" and "k" are so just plug them in at \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ 7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) and simplify and solve for "y" :)
jdoe0001
  • jdoe0001
well... negative 7 rather \(\begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ -7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\)
chrisdbest
  • chrisdbest
Ohk, yeah, I meant down 7
chrisdbest
  • chrisdbest
So what do I do next?
jdoe0001
  • jdoe0001
you shall... plug in those values simplify and solve for "y" :)
chrisdbest
  • chrisdbest
what values?
chrisdbest
  • chrisdbest
So y^2 = -28x?
chrisdbest
  • chrisdbest
@jdoe0001
jdoe0001
  • jdoe0001
\(\large \begin{array}{llll} (x-{\color{brown}{ h}})^2=4{\color{purple}{ p}}(y-{\color{blue}{ k}})\\ \end{array} \qquad \begin{array}{llll} vertex\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\\ {\color{purple}{ -7}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\) notice.... you have all the values you need, just plug them in and solve for "y"
chrisdbest
  • chrisdbest
so what's h?
jdoe0001
  • jdoe0001
look at the colors, h,k are just the vertex coordinates
chrisdbest
  • chrisdbest
ok
chrisdbest
  • chrisdbest
Then what does x equal?
chrisdbest
  • chrisdbest
Ohhh. wait I think I see
chrisdbest
  • chrisdbest
My answer choices are... A. \[y = -\frac{ 1 }{ 7 }x^2\] B. \[y^2 = -7x\] C. \[y=-\frac{ 1 }{ 28 }x^2\] D. \[y^2 = -28x\]
chrisdbest
  • chrisdbest
So I would choose D?
chrisdbest
  • chrisdbest
OMG!!! It's not D!!!
dumbcow
  • dumbcow
Answer is C note the parabola opens down, so its of the form y = x^2 the "y" is squared only if parabola is sideways or opens left/right

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