HIPERBOLA-PERSAMAAN GARIS SINGGUNG (IRISAN KERUCUT)

 Perhatikan tabel berikut

$\begin{array}{|l|l|l|}\hline \begin{aligned}&\textrm{Persamaan}\\&\textrm{hiperbola}\end{aligned}&\qquad \displaystyle \frac{x^{\displaystyle 2}}{a^{\displaystyle 2}}-\displaystyle \frac{y^{\displaystyle 2}}{b^{\displaystyle 2}}=1&\qquad \displaystyle \frac{(x-h)^{\displaystyle 2}}{a^{\displaystyle 2}}-\displaystyle \frac{(y-k)^{\displaystyle 2}}{b^{\displaystyle 2}}=1\\\hline   \begin{aligned}&\textrm{dengan}\\ &\textrm{gradien}\;\: m\end{aligned}&y=mx \pm \sqrt{a^{\displaystyle 2}m^{\displaystyle 2}-b^{\displaystyle 2}}&y-k=m(x-h)\pm \sqrt{a^{\displaystyle 2}m^{\displaystyle 2}-b^{\displaystyle 2}}\\\hline \begin{aligned}&\textrm{di titik}\\ &\left( x_{1},y_{1} \right)\\ &\textrm{pada }\\ &\textrm{hiperbola} \end{aligned}&\quad\displaystyle \frac{x_{1}x}{a^{\displaystyle 2}}-\frac{y_{1}y}{b^{\displaystyle 2}}=1&\displaystyle \frac{\left( x_{1}-h \right)(x-h)}{a^{\displaystyle 2}}-\frac{\left( y_{1}-k \right)\left( y-k \right)}{b^{\displaystyle 2}}=1\\\hline \begin{aligned}&\textrm{melalui}\\ &\textrm{titik}\\ &\left( x_{1},y_{1} \right)\\ &\textrm{di luar}\\ &\textrm{hiperbola}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\\end{aligned}&\begin{aligned}&\textrm{Prosesnya}:\\\\ &Pertama:\\ &\textrm{cari persamaan garis}\\ &\textrm{kutubnya (polar)}\\ &\textrm{dengan rumus pada}\\ &\textrm{baris 3 kolom 2}\\ &\textrm{di tabel ini}\\\\ &Kedua:\\ &\textrm{Potongkan garis kutub}\\ &\textrm{dengan hiperbola di titik}\\ &A\left( x_{A},y_{A} \right)\: \textrm{dan di titik}\\ &B\left( x_{B},y_{B} \right)\\\\ &Ketiga:\\ &\textrm{Persamaan garis}\\ &\textrm{singgung yang dicari}:\\ &\displaystyle \frac{x_{A}x}{a^{\displaystyle 2}}-\frac{y_{A}y}{b^{\displaystyle 2}}=1\\ & \textrm{dan}\\ &\displaystyle \frac{x_{B}x}{a^{\displaystyle 2}}-\frac{y_{B}y}{b^{\displaystyle 2}}=1 \end{aligned}&\begin{aligned}&\textrm{Prosenya kurang lebih sama}\\ &\textrm{dengan proses sebelah kiri}\\ &\textrm{tersebut}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ \end{aligned}\\\hline  \end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui persamaan hiperbola}\quad 9x^{\displaystyle 2}-16y^{\displaystyle 2}=144\\ &\textrm{Tentukan persamaan garis singgung hiperbola}\\ &\textrm{a}\quad\textrm{yang bergradien 1}\\ &\textrm{b}\quad\textrm{yang melalui titik}\:\: \left( 4\sqrt{2},3 \right)\\ &\textrm{c}\quad\textrm{yang melalui titik (5,0)}\\\\ &\textrm{Solusi}:\\ &\textrm{Perhatikan bahwa}\quad 9x^{\displaystyle 2}-16y^{\displaystyle 2}=144\\ &\displaystyle \frac{x^{\displaystyle 2}}{16}-\frac{y^{\displaystyle 2}}{9}=1\\ &\begin{aligned}&\bullet \quad a^{\displaystyle 2}=16\Longrightarrow a=4\\ &\bullet \quad b^{\displaystyle 2}=9\Longrightarrow b=3\\ &\bullet \quad c^{\displaystyle 2}=a^{\displaystyle 2}+b^{\displaystyle 2}=16+9=25\Longrightarrow c=5\\ &\textrm{Selanjutnya}\\ &\textrm{(a)}\quad \textrm{persamaan garis singgung dengan gradien}\:\: m=1\\ &\:\qquad y=mx\pm \sqrt{a^{\displaystyle 2}m^{\displaystyle 2}-b^{\displaystyle 2}}=1.x\pm \sqrt{16.1-9}\\ &\:\qquad \Leftrightarrow y=x\pm \sqrt{7}\\ &\textrm{(b)}\quad \textrm{persamaan garis singgung melalui titik}\:\: (4\sqrt{2},3)\\ &\:\quad \quad \textrm{dan cukup jelas bahwa titik}\:\: (4\sqrt{2},3)\:\: \textrm{pada hiperbola}\\ &\:\quad \quad \textrm{Sehingga kita dapat gunakan rumus berikut:}\\&\:\qquad \displaystyle \frac{x_{1}x}{16}-\displaystyle \frac{y_{1}y}{9}=1\Rightarrow \displaystyle \frac{4\sqrt{2}.x}{16}+\displaystyle \frac{3.y}{9}=1\\ &\:\qquad  \Leftrightarrow  3\sqrt{2}x-4y=12\\ &\textrm{(c)}\quad \textrm{persamaan garis singgung melalui titik}\:\: (3,0)\\ &\:\quad \quad \textrm{dan cukup jelas bahwa titik (3,0) ini di luar hiperbola}\\ &\:\quad \quad \textrm{Sehingga kita dapat gunakan rumus berikut:}\\ &\:\qquad \bullet\:\: \textrm{Persamaan garis kutub(polar)}\\ &\quad\qquad\displaystyle \frac{x_{1}x}{16}-\displaystyle \frac{y_{1}y}{9}=1\Rightarrow \displaystyle \frac{3x}{16}-\displaystyle \frac{0.y}{9}=1\\ &\quad\qquad  \Leftrightarrow  \displaystyle \frac{3x}{16}=1\Leftrightarrow 3x=16\Leftrightarrow x=\displaystyle \frac{16}{3}\\ &\:\qquad \bullet\:\: \textrm{Perpotongan garis kutub dengan hiperbola}\\ &\quad\qquad\displaystyle x=\displaystyle \frac{16}{3}\Rightarrow 9x^{\displaystyle 2}-16y^{\displaystyle 2}=144\\ &\quad\qquad \textrm{Alternatif 1}:\\&\quad\qquad 9\left( \displaystyle \frac{16}{3} \right)^{\displaystyle 2}-16y^{\displaystyle 2}=144\Rightarrow y=\pm \sqrt{7}\\ &\quad\qquad \textrm{selanjutnya didapat titik potongnya di}\\ &\quad\qquad \textrm{titik}\quad \left( \displaystyle \frac{16}{3},\sqrt{7} \right)\quad \textrm{dan}\quad \left( \displaystyle \frac{16}{3},\displaystyle -\sqrt{7} \right)\\ &\quad\qquad \textrm{sehingga kita tentukan garis singgungnya}\\ &\quad\qquad \textrm{dengan rumus}\quad \displaystyle \frac{x_{1}x}{16}-\displaystyle \frac{y_{1}y}{9}=1\\ &\quad\qquad \textrm{dan nantinya kita akan mendapatkan }\\ &\quad\qquad \textrm{ dua garis, yaitu}\\ &\quad\qquad y=\displaystyle -\frac{3\sqrt{7}}{7}(x-3)\quad\textrm{dan}\quad y=\displaystyle \frac{3\sqrt{7}}{7}(x-3)\\ &\quad\qquad \textrm{Alternatif 2}:\\ &\quad\qquad \textrm{Gunakan titik (3,0) dan substitusikan ke}\\  &\quad\qquad \textrm{garis}\quad y=m(x-p)+q\\ &\quad\qquad \textrm{Selanjutnya diserahkan ke pembaca}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Buatlah sketsa gambar dari soal nomor 1}\\ &\textrm{di atas}\\\\ &\textbf{Solusi}\\ &\textrm{Perhatikan sketsa gambar berikut} \end{array}$.

DAFTAR PUSTAKA

1. Muklis, Rifai, R.A. 2024. Buku Interaktif Matematika XII Tingkat Lanjut. Yogyakarta: INTAN PARIWARA EDUKASI
2. Tim MGMP Matematika. .......... Matematika IPA Kelas XII untuk SMA dan MA. .....: KARYA PUSTAKA.

HIPERBOLA (IRISAN KERUCUT)

 A. Definisi

Hiperbola adalah tempat kedudukan (lokus) titik-titik pada bidang yang memiliki selisih mutlak jarak terhadap dua titik tetap yang disebut fokus bernilai konstan

B. Persamaan hiperbola pusat O(0,0)

Perhatikan ilustrasi berikut

$\begin{aligned}&\text{Keterangan}\\ &\bullet \quad \textrm{F}_{1}\quad \textrm{dan}\quad \textrm{F}_{2}\quad \textrm{disebut fokus}\\ &\bullet \quad \textrm{A dan B disebut puncak}\\  &\bullet \quad \textrm{O disebut puncak}\\ &\bullet \quad \textrm{sumbu X sebagai sumbu utama/nyata/transversal}\\ &\qquad (\textrm{sumbu yang}\quad \textrm{F}_{1}\quad \textrm{dan}\quad \textrm{F}_{2}\quad \textrm{terletak})\\ &\bullet \quad \textrm{sumbu Y sebagai sumbu sekawan/imajiner}\\ &\bullet \quad \textrm{sumbu mayor}=AB=2a\\ &\bullet \quad \textrm{sumbu minor}=RQ=2b\\ &\bullet \quad \textrm{garis}\quad g_{1}\quad \textrm{dan}\quad g_{2}\quad \textrm{disebut garis direktris}\\ &\bullet \quad \textrm{KL dan TS disebut latus rektum, panjangnya}=\displaystyle \frac{2b^{\displaystyle 2}}{a}\\ &\bullet \quad \textrm{garis}\quad y=\pm \displaystyle \frac{b}{a}x\quad \textrm{disebut asimtot miring}\\ &\bullet \quad \textrm{di antara}:a,b,\: \textrm{dan}\: c\quad \textrm{terdapat hubungan}\quad c^{\displaystyle 2}=a^{\displaystyle 2}+b^{\displaystyle 2}\\\end{aligned}$.

$\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Hiperbola}\end{aligned}&\displaystyle \frac{x^{\displaystyle 2}}{a^{\displaystyle 2}}-\displaystyle \frac{y^{\displaystyle 2}}{b^{\displaystyle 2}}=1&\displaystyle \frac{y^{\displaystyle 2}}{a^{\displaystyle 2}}-\displaystyle \frac{x^{\displaystyle 2}}{b^{\displaystyle 2}}=1\\\hline  \textrm{Terbuka}&\textrm{kanan-kiri}&\textrm{atas-bawah}\\\hline \textrm{Fokus}&(\pm c,0)&(0,\pm c)\\\hline \textrm{Puncak}&(\pm a,0)&(0,\pm a)\\\hline \textrm{Asimtot}&y=\pm \displaystyle \frac{b}{a}x&y=\pm \displaystyle \frac{a}{b}x\\\hline \textrm{Direktris}&x=\pm \displaystyle \frac{a^{\displaystyle 2}}{c}&y=\pm \displaystyle \frac{a^{\displaystyle 2}}{c}\\\hline\begin{aligned}&\text{Sumbu}\\ &\textrm{simetri} \end{aligned}&y=0,x=0&y=0,x=0\\\hline \begin{aligned}&\text{Latus}\\ &\textrm{rektum} \end{aligned}&\displaystyle \frac{2b^{\displaystyle 2}}{a}&\displaystyle \frac{2b^{\displaystyle 2}}{a}\\\hline  \textrm{eksentrisitas}&\displaystyle \frac{c}{a}&\displaystyle \frac{c}{a}\\\hline\end{array}$.

C. Persamaan hiperbola pusat (h,k)

Perhatikan tabel berikut

$\begin{array}{|l|c|c|}\hline \begin{aligned}&\textrm{Hiperbola}\end{aligned}&\displaystyle \frac{(x-h)^{\displaystyle 2}}{a^{\displaystyle 2}}-\displaystyle \frac{(y-k)^{\displaystyle 2}}{b^{\displaystyle 2}}=1&\displaystyle \frac{(y-k)^{\displaystyle 2}}{a^{\displaystyle 2}}-\displaystyle \frac{(x-h)^{\displaystyle 2}}{b^{\displaystyle 2}}=1\\\hline  \textrm{Terbuka}&\textrm{kanan-kiri}&\textrm{atas-bawah}\\\hline \textrm{Fokus}&(h\pm c,k)&(h,k\pm c)\\\hline \textrm{Puncak}&(h\pm a,k)&(h,k\pm a)\\\hline \textrm{Asimtot}&y-k=\pm \displaystyle \frac{b}{a}(x-h)&y-k=\pm \displaystyle \frac{a}{b}(x-h)\\\hline \textrm{Direktris}&x=h\pm \displaystyle \frac{a^{\displaystyle 2}}{c}&y=k\pm \displaystyle \frac{a^{\displaystyle 2}}{c}\\\hline\begin{aligned}&\text{Sumbu}\\ &\textrm{simetri} \end{aligned}&y=k,x=h&y=k,x=h\\\hline \begin{aligned}&\text{Latus}\\ &\textrm{rektum} \end{aligned}&\displaystyle \frac{2b^{\displaystyle 2}}{a}&\displaystyle \frac{2b^{\displaystyle 2}}{a}\\\hline  \textrm{eksentrisitas}&\displaystyle \frac{c}{a}&\displaystyle \frac{c}{a}\\\hline\end{array}$.

$\LARGE{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukan koordinat pusat, puncak, fokus, asimtot}\\ &\textrm{eksentrisitas, direktris, sb.utama, sb. sekawan,}\\ &\textrm{sb. mayor, sb. minor, dan panjang latus rektum }\\ &\textrm{dari hiperbola}\quad 9x^{\displaystyle 2}-16y^{\displaystyle 2}=144\\\\ &\textrm{Solusi}:\\ &\textrm{Perhatikan bahwa}\quad 9x^{\displaystyle 2}-16y^{\displaystyle 2}=144\\ &\displaystyle \frac{x^{\displaystyle 2}}{16}-\frac{y^{\displaystyle 2}}{9}=1\\ &\begin{aligned}&\bullet \quad a^{\displaystyle 2}=16\Longrightarrow a=4\\ &\bullet \quad b^{\displaystyle 2}=9\Longrightarrow b=3\\ &\bullet \quad c^{\displaystyle 2}=a^{\displaystyle 2}+b^{\displaystyle 2}=16+9=25\Longrightarrow c=5\\ &\textrm{Sehingga diperoleh}\\ &\ast  \quad \textrm{koordinat pusat}:(0,0)\\ &\ast  \quad \textrm{koordinat puncak}:(\pm a,0)=(\pm 4,0)\\ &\ast  \quad \textrm{koordinat fokus}:(\pm c,0)=(\pm 5,0)\\ &\ast  \quad \textrm{asimtot}:y=\pm \displaystyle \frac{b}{a}x\Rightarrow y=\pm \displaystyle \frac{3}{4}x\\ &\ast  \quad \textrm{eksentrisitas}\quad e=\displaystyle \frac{c}{a}=\displaystyle \frac{5}{4}\\ &\ast  \quad \textrm{direktris}:x=\displaystyle \pm \frac{a^{\displaystyle 2}}{c}=\pm \frac{16}{5}\\ &\ast  \quad \textrm{sumbu utama}:y=0\\ &\ast  \quad \textrm{sumbu sekawan}:x=0\\ &\ast  \quad \textrm{sumbu mayor}:2a=2.4=8\\ &\ast  \quad \textrm{sumbu minor}:2b=2.3=6\\ &\ast  \quad \textrm{latus rektum}:\displaystyle \frac{2b^{\displaystyle 2}}{a}=\displaystyle \frac{18}{4}=\frac{9}{2}\\\end{aligned} \end{array}$.