Matematik

cirklens ligning og linjens ligning

02. december 2019 af matematikhaha - Niveau: A-niveau

hjælp

Vedhæftet fil: Capture2.PNG

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Svar #1
02. december 2019 af mathon

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Svar #2
02. december 2019 af janhaa

$c:\\ (x+1)^2+(y-3)^2=5^2 \\ l:\,\,x=2y-2$

https://www.wolframalpha.com/input/?i=plot+%28x%2B1%29%5E2%2B%28y-3%29%5E2%3D25+and+x%3D2y-2

gives:

(-4, -1) and (4; 3)

$2(x+1)+2(y-3)*y'=0\\ \\ y'=-3/4\\ \\ y(tangent)=y=-\frac{3}{4}x-4$

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Svar #3
02. december 2019 af janhaa

https://www.wolframalpha.com/input/?i=plot+%28x%2B1%29%5E2%2B%28y-3%29%5E2%3D25+and+x%3D2y-2+and+y%3D-0.75x-4

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Svar #4
02. december 2019 af mathon

$\small \begin{array}{lllll} a)&&\left ( x-1 \right )^2+\left ( y-3\right ) ^2=15+1+9=5^2\\\\ &\textup{sk\ae ringspunkter:}&\left ( 2y-2-1 \right )^2+\left ( y-3\right ) ^2=5^2\\\\ &&\left ( 2y-3) ^2+(y-3)^2\right )=25\\\\ &&4y^2-12y+9+y^2-6y+9=25\\\\ &&5y^2-18y-7=0\\\\ &&y=\left\{\begin{array}{lll}\frac{9-2\sqrt{29}}{5}\\\\\frac{9+2\sqrt{29}}{5} \end{array}\right.\\\\ &&x=\left\{\begin{array}{lll}\frac{8-4\sqrt{29}}{5}\\\\\frac{8+4\sqrt{29}}{5} \end{array}\right.\\\\ &\textup{dvs}&S_1=\left ( \frac{8-4\sqrt{29}}{5},\frac{9-2\sqrt{29}}{5} \right )\textup{ og }S_2=\left ( \frac{8+4\sqrt{29}}{5},\frac{9+2\sqrt{29}}{5} \right ) \end{array}$

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Svar #5
02. december 2019 af janhaa

https://www.wolframalpha.com/input/?i=intersection+%28x+%2B1%29%5E2%2B+%28y-3%29%5E2%3D25+and+x%3D2y-2

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Svar #6
02. december 2019 af mathon

Jeg har en tegnfejl i

$\small \begin{array}{llll} (x-1)^2+(y-3)^2=5^2&\textup{som skal v\ae re}\\\\ (x+1)^2+(y-3)^2=5^2&\textup{hvorefter det hele bliver noget enklere}\\ \textup{som i }\#2\\\\ \textup{Brug denne.} \end{array}$

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Svar #7
03. december 2019 af mathon

$\small \begin{array}{lllll} \textup{tangent i (-4,-1):}&(-4+1)(x+1)+(-1-3)(y-3)=5^2\\\\ &-3(x+1)-4(y-3)=25\\\\ &-3x-4y-3+12=25\\\\ &-3x-16=4y\\\\ &y=-\frac{3}{4}x-4 \end{array}$

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