{
"query": {
"display": "$$y^{2}+8x+10y-5=0$$",
"symbolab_question": "CONIC#y^{2}+8x+10y-5=0"
},
"solution": {
"level": "PERFORMED",
"subject": "Geometry",
"topic": "Parabola",
"subTopic": "formula",
"default": "(h,k)=(\\frac{15}{4},-5),p=-2"
},
"steps": {
"type": "interim",
"title": "$$y^{2}+8x+10y-5=0:\\quad$$Parabola with vertex at $$\\left(h,\\:k\\right)=\\left(\\frac{15}{4},\\:-5\\right),\\:$$and focal length $$|p|=2$$",
"input": "y^{2}+8x+10y-5=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 $$y^{2}+8x+10y-5=0\\:$$in the standard form",
"input": "y^{2}+8x+10y-5=0",
"steps": [
{
"type": "step",
"primary": "Rewrite as",
"result": "8x=-y^{2}-10y+5"
},
{
"type": "step",
"primary": "Divide by $$8$$",
"result": "x=-\\frac{y^{2}}{8}-\\frac{5y}{4}+\\frac{5}{8}"
},
{
"type": "interim",
"title": "Complete the square $$-\\frac{y^{2}}{8}-\\frac{5y}{4}+\\frac{5}{8}:{\\quad}-\\frac{1}{8}\\left(y+5\\right)^{2}+\\frac{15}{4}$$",
"input": "-\\frac{y^{2}}{8}-\\frac{5y}{4}+\\frac{5}{8}",
"steps": [
{
"type": "step",
"primary": "Write $$-\\frac{y^{2}}{8}-\\frac{5y}{4}+\\frac{5}{8}\\:$$in the form: $$x^2+2ax+a^2$$",
"secondary": [
"Factor out $$-\\frac{1}{8}$$"
],
"result": "-\\frac{1}{8}\\left(y^{2}+10y-5\\right)"
},
{
"type": "interim",
"title": "$$2a=10{\\quad:\\quad}a=5$$",
"input": "2a=10",
"steps": [
{
"type": "interim",
"title": "Divide both sides by $$2$$",
"input": "2a=10",
"result": "a=5",
"steps": [
{
"type": "step",
"primary": "Divide both sides by $$2$$",
"result": "\\frac{2a}{2}=\\frac{10}{2}"
},
{
"type": "step",
"primary": "Simplify",
"result": "a=5"
}
],
"meta": {
"interimType": "Divide Both Sides Specific 1Eq",
"gptData": "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"
}
}
],
"meta": {
"solvingClass": "Equations",
"interimType": "Equations"
}
},
{
"type": "step",
"primary": "Add and subtract $$5^{2}\\:$$",
"result": "-\\frac{1}{8}\\left(y^{2}+10y-5+5^{2}-5^{2}\\right)"
},
{
"type": "step",
"primary": "$$x^2+2ax+a^2=\\left(x+a\\right)^2$$",
"secondary": [
"$$y^{2}+10y+5^{2}=\\left(y+5\\right)^{2}$$",
"Complete the square"
],
"result": "-\\frac{1}{8}\\left(\\left(y+5\\right)^{2}-5-5^{2}\\right)"
},
{
"type": "step",
"primary": "Simplify",
"result": "-\\frac{1}{8}\\left(y+5\\right)^{2}+\\frac{15}{4}"
}
],
"meta": {
"solvingClass": "Equations",
"interimType": "Complete Square 1Eq"
}
},
{
"type": "step",
"result": "x=-\\frac{1}{8}\\left(y+5\\right)^{2}+\\frac{15}{4}"
},
{
"type": "step",
"primary": "Subtract $$\\frac{15}{4}$$ from both sides",
"result": "x-\\frac{15}{4}=-\\frac{1}{8}\\left(y+5\\right)^{2}"
},
{
"type": "step",
"primary": "Divide by coefficient of square terms: $$-\\frac{1}{8}$$",
"result": "-8\\left(x-\\frac{15}{4}\\right)=\\left(y+5\\right)^{2}"
},
{
"type": "step",
"primary": "Rewrite in standard form",
"result": "4\\left(-2\\right)\\left(x-\\frac{15}{4}\\right)=\\left(y-\\left(-5\\right)\\right)^{2}"
}
],
"meta": {
"interimType": "Parabola Canonical Format 1Eq"
}
},
{
"type": "step",
"result": "4\\left(-2\\right)\\left(x-\\frac{15}{4}\\right)=\\left(y-\\left(-5\\right)\\right)^{2}"
},
{
"type": "step",
"primary": "Therefore parabola properties are:",
"result": "\\left(h,\\:k\\right)=\\left(\\frac{15}{4},\\:-5\\right),\\:p=-2"
}
],
"meta": {
"solvingClass": "Parabola"
}
},
"plot_output": {
"meta": {
"plotInfo": {
"variable": "x",
"funcsToDraw": {
"funcs": [
{
"evalFormula": "y=\\sqrt{4(-2)(x-\\frac{15}{4})}-5",
"displayFormula": "4(-2)(x-\\frac{15}{4})=(y-(-5))^{2}",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
},
{
"evalFormula": "y=-\\sqrt{4(-2)(x-\\frac{15}{4})}-5",
"displayFormula": "4(-2)(x-\\frac{15}{4})=(y-(-5))^{2}",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
},
{
"evalFormula": "x=\\frac{23}{4}",
"displayFormula": "x=\\frac{23}{4}",
"attributes": {
"color": "GRAY",
"lineType": "NORMAL",
"labels": [
"\\mathrm{directrix}"
],
"isAsymptote": false
}
}
]
},
"pointsToDraw": {
"pointsLatex": [
"(\\frac{15}{4},-5)",
"(\\frac{7}{4},-5)"
],
"pointsDecimal": [
{
"fst": 3.75,
"snd": -5
},
{
"fst": 1.75,
"snd": -5
}
],
"attributes": [
{
"color": "PURPLE",
"labels": [
"\\mathrm{vertex}"
],
"labelTypes": [
"DEFAULT"
],
"labelColors": [
"PURPLE"
]
},
{
"color": "PURPLE",
"labels": [
"\\mathrm{focus}"
],
"labelTypes": [
"DEFAULT"
],
"labelColors": [
"PURPLE"
]
}
]
},
"functionChanges": [
{
"origFormulaLatex": [],
"finalFormulaLatex": [],
"plotTitle": "4(-2)(x-\\frac{15}{4})=(y-(-5))^{2}",
"paramsLatex": [],
"paramsReplacementsLatex": []
}
],
"localBoundingBox": {
"xMin": -18.75,
"xMax": 26.25,
"yMin": -27.5,
"yMax": 17.5
}
},
"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+8x+10y-5=0 ?
The solution to y^2+8x+10y-5=0 is Parabola with (h,k)=(15/4 ,-5),p=-2