{
"query": {
"display": "$$9x^{2}-y^{2}-36x-6y+18=0$$",
"symbolab_question": "CONIC#9x^{2}-y^{2}-36x-6y+18=0"
},
"solution": {
"level": "PERFORMED",
"subject": "Geometry",
"topic": "Hyperbola",
"subTopic": "formula",
"default": "(h,k)=(2,-3),a=1,b=3"
},
"steps": {
"type": "interim",
"title": "$$9x^{2}-y^{2}-36x-6y+18=0:\\quad$$Hipérbola que abre hacia la derecha y hacia la izquierda $$\\left(h,\\:k\\right)=\\left(2,\\:-3\\right),\\:a=1,\\:b=3$$",
"input": "9x^{2}-y^{2}-36x-6y+18=0",
"steps": [
{
"type": "definition",
"title": "Ecuación normal de la hipérbola",
"text": "$$\\frac{\\left(x-h\\right)^{2}}}{a^2}}}-\\frac{\\left(y-k\\right)^{2}}}{b^2}}}=1\\mathrm{\\:es\\:la\\:ecuación\\:estándar\\:para\\:una\\:hipérbola\\:orientada\\:de\\:derecha-izquierda.}$$<br/>Con centro $$\\bold{\\left(h,\\:k\\right)},\\:$$ semieje $$\\bold{a}$$ y semieje conjugado $$\\bold{b}$$."
},
{
"type": "interim",
"title": "Reescribir $$9x^{2}-y^{2}-36x-6y+18=0\\:$$con la forma de la ecuación general de la hipérbola",
"input": "9x^{2}-y^{2}-36x-6y+18=0",
"steps": [
{
"type": "step",
"primary": "Restar $$18$$ de ambos lados",
"result": "9x^{2}-36x-y^{2}-6y=-18"
},
{
"type": "step",
"primary": "Factorizar el coeficiente de términos cuadrados",
"result": "9\\left(x^{2}-4x\\right)-\\left(y^{2}+6y\\right)=-18"
},
{
"type": "step",
"primary": "Dividir entre el coeficiente de términos cuadrados: $$9$$",
"result": "\\left(x^{2}-4x\\right)-\\frac{1}{9}\\left(y^{2}+6y\\right)=-2"
},
{
"type": "step",
"primary": "Dividir entre el coeficiente de términos cuadrados: $$1$$",
"result": "\\frac{1}{1}\\left(x^{2}-4x\\right)-\\frac{1}{9}\\left(y^{2}+6y\\right)=-2"
},
{
"type": "step",
"primary": "Convertir $$x\\:$$a su forma cuadrática",
"result": "\\frac{1}{1}\\left(x^{2}-4x+4\\right)-\\frac{1}{9}\\left(y^{2}+6y\\right)=-2+\\frac{1}{1}\\left(4\\right)"
},
{
"type": "step",
"primary": "Convertir a forma cuadrática",
"result": "\\frac{1}{1}\\left(x-2\\right)^{2}-\\frac{1}{9}\\left(y^{2}+6y\\right)=-2+\\frac{1}{1}\\left(4\\right)"
},
{
"type": "step",
"primary": "Convertir $$y\\:$$a su forma cuadrática",
"result": "\\frac{1}{1}\\left(x-2\\right)^{2}-\\frac{1}{9}\\left(y^{2}+6y+9\\right)=-2+\\frac{1}{1}\\left(4\\right)-\\frac{1}{9}\\left(9\\right)"
},
{
"type": "step",
"primary": "Convertir a forma cuadrática",
"result": "\\frac{1}{1}\\left(x-2\\right)^{2}-\\frac{1}{9}\\left(y+3\\right)^{2}=-2+\\frac{1}{1}\\left(4\\right)-\\frac{1}{9}\\left(9\\right)"
},
{
"type": "step",
"primary": "Simplificar $$-2+\\frac{1}{1}\\left(4\\right)-\\frac{1}{9}\\left(9\\right)$$",
"result": "\\frac{1}{1}\\left(x-2\\right)^{2}-\\frac{1}{9}\\left(y+3\\right)^{2}=1"
},
{
"type": "step",
"primary": "Simplificar",
"result": "\\frac{\\left(x-2\\right)^{2}}{1}-\\frac{\\left(y+3\\right)^{2}}{9}=1"
},
{
"type": "step",
"primary": "Reescribir en la forma estándar",
"result": "\\frac{\\left(x-2\\right)^{2}}{1^{2}}-\\frac{\\left(y-\\left(-3\\right)\\right)^{2}}{3^{2}}=1"
}
],
"meta": {
"interimType": "Hyperbola Canonical Format 1Eq"
}
},
{
"type": "step",
"result": "\\frac{\\left(x-2\\right)^{2}}{1^{2}}-\\frac{\\left(y-\\left(-3\\right)\\right)^{2}}{3^{2}}=1"
},
{
"type": "step",
"primary": "Por lo tanto, las propiedades de la hipérbola son: ",
"result": "\\left(h,\\:k\\right)=\\left(2,\\:-3\\right),\\:a=1,\\:b=3"
}
],
"meta": {
"solvingClass": "Hyperbola"
}
},
"plot_output": {
"meta": {
"plotInfo": {
"variable": "x",
"funcsToDraw": {
"funcs": [
{
"evalFormula": "y=3(x-2)-3",
"displayFormula": "y=3(x-2)-3",
"attributes": {
"color": "PURPLE",
"lineType": "DASH",
"isAsymptote": true
}
},
{
"evalFormula": "y=-3(x-2)-3",
"displayFormula": "y=-3(x-2)-3",
"attributes": {
"color": "PURPLE",
"lineType": "DASH",
"isAsymptote": true
}
},
{
"evalFormula": "y=\\sqrt{9(\\frac{(x-2)^{2}}{1^{2}}-1)}-3",
"displayFormula": "\\frac{(x-2)^{2}}{1^{2}}-\\frac{(y-(-3))^{2}}{3^{2}}=1",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
},
{
"evalFormula": "y=-\\sqrt{9(\\frac{(x-2)^{2}}{1^{2}}-1)}-3",
"displayFormula": "\\frac{(x-2)^{2}}{1^{2}}-\\frac{(y-(-3))^{2}}{3^{2}}=1",
"attributes": {
"color": "PURPLE",
"lineType": "NORMAL",
"isAsymptote": false
}
}
]
},
"pointsToDraw": {
"pointsLatex": [
"(2,-3)"
],
"pointsDecimal": [
{
"fst": 2,
"snd": -3
}
],
"attributes": [
{
"color": "PURPLE",
"labels": [
"\\mathrm{Center}"
],
"labelTypes": [
"DEFAULT"
],
"labelColors": [
"PURPLE"
]
}
]
},
"linesToDraw": [
{
"p1x": "2",
"p1y": "-3",
"p2x": "3",
"p2y": "-3",
"attributes": {
"color": "GRAY",
"lineType": "BOLD",
"labels": [
"a=1"
],
"isAsymptote": false
}
},
{
"p1x": "2",
"p1y": "-3",
"p2x": "2",
"p2y": "0",
"attributes": {
"color": "GRAY",
"lineType": "BOLD",
"labels": [
"b=3"
],
"isAsymptote": false
}
}
],
"functionChanges": [
{
"origFormulaLatex": [],
"finalFormulaLatex": [],
"plotTitle": "\\frac{(x-2)^{2}}{1^{2}}-\\frac{(y-(-3))^{2}}{3^{2}}=1",
"paramsLatex": [],
"paramsReplacementsLatex": []
}
],
"localBoundingBox": {
"xMin": -12.214285714285714,
"xMax": 14.785714285714286,
"yMin": -16.02857142857143,
"yMax": 10.971428571428572
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},
"showViewLarger": true
}
}
}
Solución
Solución
Pasos de solución
Reescribir con la forma de la ecuación general de la hipérbola
Por lo tanto, las propiedades de la hipérbola son: