CONTOH SOAL 3 IRISAN KERUCUT

 $\begin{array}{ll}\\ 5.&\textbf{KSM MA 2013 Tk. KABUPATEN}\\ &\textrm{Diberikan titik} \quad A\left( -3,9\right)\quad \textrm{dan titik}\quad B\left( x,x^{\displaystyle 2} \right)\\ &\textrm{dengan}\quad x\neq 0\quad \textrm{pada parabola}\quad y=x^{\displaystyle 2}\\ &\textrm{sehingga terdapat lingkaran L yang melalui}\\ &\textrm{titik A, B, dan O}(0,0)\\   \end{array}$.

$\begin{array}{ll}\\ .\:\quad&\textrm{Jika ruas garis}\:\:\overline{AB}\:\: \textrm{adalah diameter lingkaran}\\ &\textrm{tersebut, maka koordinat titik B adalah} ...\\ &\begin{array}{llll} \textrm{A}.&\displaystyle (1,1)\\ \textrm{B}.&\displaystyle \left( \displaystyle \frac{1}{2},\frac{1}{4} \right)\\ \textrm{C}.&\displaystyle \left( \displaystyle \frac{1}{3},\frac{1}{9} \right)\\ \textrm{D}.&\displaystyle \left( \displaystyle \frac{2}{3},\frac{4}{9} \right)\\ \textrm{E}.&\displaystyle \left( \displaystyle \frac{2}{5},\frac{4}{25} \right)\end{array}\\\\ &\textbf{Jawab}:\quad  \textbf{C}\\ &\begin{aligned}&\text{Diketahui bahwa}\:\: AB\:\: \textrm{adalah diameter lingkaran}\\ &\text{dan O terletak pada lingkaran, maka}\\ &\angle AOB=90^{0}.\quad \textrm{Akibatnya vektor}\:\: \overrightarrow{OA}=(-3,9)\\ &\textrm{dan vektor}\quad \overrightarrow{OB}=\left( x,x^{\displaystyle 2} \right)\:\: \textrm{saling tegak lurus}.\\ &\textrm{Secara dot product}:\\ &\overrightarrow{OA}.\overrightarrow{OB}=0\Leftrightarrow (-3)(x)+(9)(x^{2})=0\\ &\Leftrightarrow 3x(3x-1)=0\Leftrightarrow x=\displaystyle \frac{1}{3}\Rightarrow x^{\displaystyle 2}=\displaystyle \frac{1}{9}\\ &\textrm{Sehingga}\quad \left( x,x^{\displaystyle 2} \right)=\left( \displaystyle \frac{1}{3},\frac{1}{9} \right) \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 6.&\textbf{EBTANAS SMA IPA 1996}\\ &\textrm{Hiperbola yang berfokus di titik} \quad \left( \displaystyle 5,0 \right)\\ & \textrm{berpusat di titik}\quad (0,0)\quad \textrm{dan panjang sumbu}\\ &\textrm{mayor}=8,\quad \textrm{persamaannya adalah}\:....\\ &\begin{array}{llll}\\ \textrm{A}.&\displaystyle \frac{x^{2}}{64}-\frac{y^{2}}{36}=1\\ \textrm{B}.&\displaystyle \frac{x^{2}}{25}-\frac{y^{2}}{16}=1\\ \textrm{C}.&\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1\\ \textrm{D}.&\displaystyle \frac{y^{2}}{25}-\frac{x^{2}}{9}=1\\ \textrm{E}.&\displaystyle \frac{y^{2}}{16}-\frac{x^{2}}{9}=1\end{array}\\\\ &\textrm{Jawab}:\quad \textbf{C}\\ &\textrm{Perhatikan bahwa hiperbola dengan pusat}:\\ &(0,0)\:\: \textrm{dan berfokus di}\:\: (5,0)\Rightarrow c=5\quad \textrm{akan }\\ &\textrm{memiliki persamaan hiperbola}:\displaystyle \frac{x^{\displaystyle 2}}{a^{\displaystyle 2}}-\frac{y^{\displaystyle 2}}{b^{\displaystyle 2}}=1\\ &\begin{aligned}&\textrm{dan untuk panjang sb.mayor(sb. transversal)}=8,\\ &\textrm{artinya}: 2a=8\Leftrightarrow a=4.\:\: \textrm{Sedangkan untuk}\\ &b=\sqrt{c^{\displaystyle 2}-a^{\displaystyle 2}}=\sqrt{5^{\displaystyle 2}-4^{\displaystyle 2}}=\sqrt{25-16}=\sqrt{9}=3\\ &\textrm{Sehingga persamaan hiperbolanya adalah}:\\ &\displaystyle \frac{x^{2}}{4^{\displaystyle 2}}-\frac{y^{2}}{3^{\displaystyle 2}}=1\Leftrightarrow \displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 \end{aligned}   \end{array}$.