Matematika otak saya ngelag nyari ini, ada yg bisa bantu?​

mathildemarquardtulv – March 07, 2023

otak saya ngelag nyari ini, ada yg bisa bantu?​

Kurva dengan persamaan [tex]xy=1[/tex] dirotasikan terhadap pusat [tex]P(2, 6)[/tex] sebesar [tex]135^{\circ}[/tex] searah jarum jam persamaannya

[tex]\big(-x+y-4+2\sqrt2\big)\big(-x-y+8+6\sqrt2\big)=2[/tex]

Penyelesaian

pertama kita gunakan matriks transformasi rotasi

untuk rotasi terpusat pada [tex](0, 0)[/tex] matriksnya

[tex]\begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \end{pmatrix}\begin{pmatrix}x \\y \end{pmatrix}[/tex]

untuk rotasi dengan pusat [tex](a,b)[/tex] matriksnya

[tex]\begin{pmatrix}x' -a\\y' -b\end{pmatrix}=\begin{pmatrix}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \end{pmatrix}\begin{pmatrix}x -a\\y -b \end{pmatrix}[/tex]

atau

[tex]\begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}\cos{\theta} & -\sin{\theta} \\\sin{\theta} & \cos{\theta} \end{pmatrix}\begin{pmatrix}x -a\\y -b \end{pmatrix}+\begin{pmatrix}a\\b \end{pmatrix}[/tex]

diketahui pusatnya [tex]P=(2,6)[/tex] dan perputaran rotasinya [tex]135^{\circ}[/tex] searah jarum jam.
Karena searah jarum jam maka sudutnya negatif sehingga

[tex]\begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}\cos{(-135^{\circ})} & -\sin{(-135^{\circ})} \\\sin{(-135^{\circ})} & \cos{(-135^{\circ})} \end{pmatrix}\begin{pmatrix}x -2\\y -6 \end{pmatrix}+\begin{pmatrix}2\\6 \end{pmatrix}[/tex]

kemudian kita cari nilai [tex]\cos(-135^{\circ})[/tex] dan [tex]\sin(-135^{\circ})[/tex] untuk menyederhanakan matriks tersebut

[tex]\cos(-135^{\circ})=\cos(360-135)^{\circ}[/tex]

[tex]=\cos225^{\circ}[/tex]

[tex]=\cos(180^{\circ}+45^{\circ})[/tex]

[tex]=-\cos45^{\circ}[/tex]

[tex]\Rightarrow -\displaystyle{\frac{\sqrt{2}}{2}}[/tex]

[tex]\sin(-135^{\circ})=-\sin135^{\circ}[/tex]

[tex]=-\sin(180^{\circ}-45^{\circ})[/tex]

[tex]=-\sin45^{\circ}[/tex]

[tex]\Rightarrow -\displaystyle{\frac{\sqrt{2}}{2}}[/tex]

sederhanakan matriksnya

[tex]\begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}\cos{(-135^{\circ})} & -\sin{(-135^{\circ})} \\\sin{(-135^{\circ})} & \cos{(-135^{\circ})} \end{pmatrix}\begin{pmatrix}x -2\\y -6 \end{pmatrix}+\begin{pmatrix}2\\6 \end{pmatrix}[/tex]

[tex]\Rightarrow \begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}-\displaystyle{\frac{\sqrt{2}}{2}} & \displaystyle{\frac{\sqrt{2}}{2}} \\-\displaystyle{\frac{\sqrt{2}}{2}} & -\displaystyle{\frac{\sqrt{2}}{2}} \end{pmatrix}\begin{pmatrix}x -2\\y -6 \end{pmatrix}+\begin{pmatrix}2\\6 \end{pmatrix}[/tex]

[tex]\Rightarrow \begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}-\displaystyle{\frac{\sqrt{2}}{2}}(x-2)+\displaystyle{\frac{\sqrt{2}}{2}}(y-6) \\-\displaystyle{\frac{\sqrt{2}}{2}}(x-2)-\displaystyle{\frac{\sqrt{2}}{2}}(y-6) \end{pmatrix}+\begin{pmatrix}2\\6 \end{pmatrix}[/tex]

[tex]\Rightarrow \begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}-\displaystyle{\frac{\sqrt{2}}{2}}(x-2)+\displaystyle{\frac{\sqrt{2}}{2}}(y-6) +2\\-\displaystyle{\frac{\sqrt{2}}{2}}(x-2)-\displaystyle{\frac{\sqrt{2}}{2}}(y-6) +6\end{pmatrix}[/tex]

[tex]\Rightarrow \begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}-\displaystyle{\frac{\sqrt{2}}{2}}x+\sqrt2+\displaystyle{\frac{\sqrt{2}}{2}}y-3\sqrt2 +2\\-\displaystyle{\frac{\sqrt{2}}{2}}x+\sqrt2-\displaystyle{\frac{\sqrt{2}}{2}}y+3\sqrt2 +6\end{pmatrix}[/tex]

[tex]\Rightarrow \begin{pmatrix}x' \\y' \end{pmatrix}=\begin{pmatrix}-\displaystyle{\frac{\sqrt{2}}{2}}x+\displaystyle{\frac{\sqrt{2}}{2}}y-2\sqrt2 +2\\-\displaystyle{\frac{\sqrt{2}}{2}}x-\displaystyle{\frac{\sqrt{2}}{2}}y+4\sqrt2 +6\end{pmatrix}[/tex]

sehingga transformansinya

ubah [tex]x[/tex] menjadi [tex]-\displaystyle{\frac{\sqrt{2}}{2}}x+\displaystyle{\frac{\sqrt{2}}{2}}y-2\sqrt2 +2[/tex]

ubah [tex]y[/tex] menjadi [tex]-\displaystyle{\frac{\sqrt{2}}{2}}x-\displaystyle{\frac{\sqrt{2}}{2}}y+4\sqrt2 +6[/tex]

untuk kurva [tex]xy=1[/tex] yang dirotasikan terhadap pusat [tex]P(2,6)[/tex] sebesar [tex]135^{\circ}[/tex] SJJ, persamaan kurvanya

[tex]\bigg(-\displaystyle{\frac{\sqrt2}{2}}x+\displaystyle{\frac{\sqrt2}{2}}y-2\sqrt2+2\bigg)\bigg(-\displaystyle{\frac{\sqrt2}{2}}x-\displaystyle{\frac{\sqrt2}{2}}y+4\sqrt2+6\bigg)=1[/tex]

[tex]\Rightarrow\displaystyle{\frac{\sqrt2}{2}}\bigg(-x+y-4+2\sqrt2\bigg)\displaystyle{\frac{\sqrt2}{2}}\bigg(-x-y+8+6\sqrt2\bigg)=1[/tex]

[tex]\Rightarrow\displaystyle{\frac{2}{4}}\big(-x+y-4+2\sqrt2\big)\big(-x-y+8+6\sqrt2\big)=1[/tex]

[tex]\Rightarrow\displaystyle{\frac{1}{2}}\big(-x+y-4+2\sqrt2\big)\big(-x-y+8+6\sqrt2\big)=1[/tex]

atau

[tex]\Rightarrow\big(-x+y-4+2\sqrt2\big)\big(-x-y+8+6\sqrt2\big)=2[/tex]

nb : jika rotasinya [tex]135^{\circ}[/tex] berlawanan arah jarum jam, maka kurvanya

[tex]\bigg(\displaystyle{\frac{\sqrt2}{2}}x-\displaystyle{\frac{\sqrt2}{2}}y+2\sqrt2+6\bigg)\bigg(-\displaystyle{\frac{\sqrt2}{2}}x-\displaystyle{\frac{\sqrt2}{2}}y+4\sqrt2+2\bigg)=1[/tex]

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