{
"query": {
"display": "$$y^{2}+4x+1=0$$",
"symbolab_question": "CONIC#y^{2}+4x+1=0"
},
"solution": {
"level": "PERFORMED",
"subject": "Geometry",
"topic": "Parabola",
"subTopic": "formula",
"default": "(h,k)=(-\\frac{1}{4},0),p=-1"
},
"steps": {
"type": "interim",
"title": "$$y^{2}+4x+1=0:\\quad$$Parabola with vertex at $$\\left(h,\\:k\\right)=\\left(-\\frac{1}{4},\\:0\\right),\\:$$and focal length $$|p|=1$$",
"input": "y^{2}+4x+1=0",
"steps": [
{
"type": "definition",
"title": "Parabola standard equation",
"text": "$$4p\\left(x-h\\right)=\\left(y-k\\right)^{2}\\:$$ is the standard equation for a right-left facing parabola with vertex at $$\\left(h,\\:k\\right),\\:$$<br/>and a focal length $$|p|$$"
},
{
"type": "interim",
"title": "Rewrite $$y^{2}+4x+1=0\\:$$in the standard form:$${\\quad}4\\left(-1\\right)\\left(x-\\left(-\\frac{1}{4}\\right)\\right)=\\left(y-0\\right)^{2}$$",
"input": "y^{2}+4x+1=0",
"steps": [
{
"type": "step",
"primary": "Subtract $$y^{2}$$ from both sides",
"result": "y^{2}+4x+1-y^{2}=0-y^{2}"
},
{
"type": "step",
"primary": "Refine",
"result": "4x+1=-y^{2}"
},
{
"type": "step",
"primary": "Divide both sides by $$-1$$",
"result": "\\frac{4x+1}{-1}=\\frac{-y^{2}}{-1}"
},
{
"type": "step",
"primary": "Simplify",
"result": "-4x-1=y^{2}"
},
{
"type": "step",
"primary": "Factor $$-4$$",
"result": "\\left(-4\\right)\\left(x+\\frac{-1}{-4}\\right)=y^{2}"
},
{
"type": "step",
"primary": "Simplify",
"result": "\\left(-4\\right)\\left(x+\\frac{1}{4}\\right)=y^{2}"
},
{
"type": "step",
"primary": "Factor $$4$$",
"result": "4\\cdot\\:\\frac{-4}{4}\\left(x+\\frac{1}{4}\\right)=y^{2}"
},
{
"type": "step",
"primary": "Simplify",
"result": "4\\left(-1\\right)\\left(x+\\frac{1}{4}\\right)=y^{2}"
},
{
"type": "step",
"primary": "Rewrite as",
"result": "4\\left(-1\\right)\\left(x-\\left(-\\frac{1}{4}\\right)\\right)=\\left(y-0\\right)^{2}"
}
],
"meta": {
"interimType": "Parabola Canonical Format Title 1Eq",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7aRgRQFW0MqhVT4GbzWEc4smfnf+f7rxYzWg3st7WK0axzzcMHjSLmak70ycPvkgfLHlRs0F2lS8z71EJhfVfEQcoiLKrONfIA8+fB2d3wo49eajA7BpGfE9oODIxje87sUcw1ttPAkiUYG2MHwFsdafud2BBSI5iFqZd6pwGMxpN5Aod6Hr1Lp2e/29KhSgU7CCRPNHKioRGFeNUeDwBhP/ilKgkmpRap8RWThRhtKSB64Ml8b0Ki0WaGopRmMsmH0/Zy5mINpsQVQVsbT3LEg=="
}
},
{
"type": "step",
"primary": "Therefore parabola properties are:",
"result": "\\left(h,\\:k\\right)=\\left(-\\frac{1}{4},\\:0\\right),\\:p=-1"
}
],
"meta": {
"solvingClass": "Parabola"
}
},
"plot_output": {
"meta": {
"plotInfo": {
"variable": "x",
"funcsToDraw": {
"funcs": [
{
"evalFormula": "y=\\sqrt{4(-1)(x-(-\\frac{1}{4}))}+0",
"displayFormula": "4(-1)(x-(-\\frac{1}{4}))=y^{2}",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
},
{
"evalFormula": "y=-\\sqrt{4(-1)(x-(-\\frac{1}{4}))}+0",
"displayFormula": "4(-1)(x-(-\\frac{1}{4}))=y^{2}",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
},
{
"evalFormula": "x=\\frac{3}{4}",
"displayFormula": "x=\\frac{3}{4}",
"attributes": {
"color": "GRAY",
"lineType": "NORMAL",
"labels": [
"\\mathrm{directrix}"
],
"isAsymptote": false
}
}
]
},
"pointsToDraw": {
"pointsLatex": [
"(-\\frac{1}{4},0)",
"(-\\frac{5}{4},0)"
],
"pointsDecimal": [
{
"fst": -0.25,
"snd": 0
},
{
"fst": -1.25,
"snd": 0
}
],
"attributes": [
{
"color": "PURPLE",
"labels": [
"\\mathrm{vertex}"
],
"labelTypes": [
"DEFAULT"
],
"labelColors": [
"PURPLE"
]
},
{
"color": "PURPLE",
"labels": [
"\\mathrm{focus}"
],
"labelTypes": [
"DEFAULT"
],
"labelColors": [
"PURPLE"
]
}
]
},
"functionChanges": [
{
"origFormulaLatex": [],
"finalFormulaLatex": [],
"plotTitle": "4(-1)(x-(-\\frac{1}{4}))=y^{2}",
"paramsLatex": [],
"paramsReplacementsLatex": []
}
],
"localBoundingBox": {
"xMin": -11.5,
"xMax": 11,
"yMin": -11.25,
"yMax": 11.25
}
},
"showViewLarger": true
}
}
}
Solution
Solution
Solution steps
Rewrite in the standard form:
Therefore parabola properties are:
Graph
Popular Examples
Frequently Asked Questions (FAQ)
What is y^2+4x+1=0 ?
The solution to y^2+4x+1=0 is Parabola with (h,k)=(-1/4 ,0),p=-1