4.4-bo'lim : Giperbolalar
4. Tenglamaning \(x\) va \(y\) qismlaridagi kvadratni to‘ldiring va tenglamani giperbola tenglamasining standart ko‘rinishiga yozing.
\[4{x^2} - 32x - {y^2} - 4y + 24 = 0\]Yechim
Bu erda bo'ladigan jarayon biz oldingi bo'limda ellips tenglamalarini standart shaklda yozish uchun ishlatgan jarayon bilan bir xil bo'ladi.
Birinchi qadam \({x^2}\) va \({y^2}\) koeffitsientlari bitta ekanligiga ishonch hosil qilishdir. \({x^2}\) koeffitsienti 4, \({y^2}\) esa -1 koeffitsientiga ega. Biz qiladigan narsa \(x\) bilan bog'liq har bir atamadan 4 ni va \(y\) bilan bog'liq har doimgidan -1 ni ko'paytiramiz. Shunday qilib,
\[4\left( {{x^2} - 8x} \right) - \left( {{y^2} + 4y} \right) + 24 = 0\]Bunday muammolardan ehtiyot bo'ling va hatto -1 koeffitsientiga ham e'tibor berish kerakligini unutmang!
2-qadam
Endi kvadratni to'ldirishni boshlaylik. Birinchidan, bizga \(x\) va \(y\) atamalarining yarmi koeffitsienti kerak, har birining kvadrati va bu raqamlarni tegishli joylarda quyidagi tarzda qo'shish/ayirish,
\[{\left( {\frac{{ - 8}}{2}} \right)^2} = {\left( { - 4} \right)^2} = \,{\color{Red} 16}\hspace{0.25in}{\left( {\frac{4}{2}} \right)^2} = {\left( 2 \right)^2} = \,{\color{ProcessBlue} 4}\] \[4\left( {{x^2} - 8x \,{\color{Red} + 16 - 16}} \right) - \left( {{y^2} + 4y \,{\color{ProcessBlue} + 4 - 4}} \right) + 24 = 0\]Bu raqamlarni qo'shish/ayirishda ehtiyot bo'ling va ularni qavs ichiga qo'yganingizga ishonch hosil qiling!
3-qadam
Keyinchalik, biz \(x\) va \(y\) shartlarini faktorga kiritishimiz va barcha konstantalarni qo'shishimiz kerak.
\[\begin{align*}4\left( {{{\left( {x - 4} \right)}^2} \,{\color{Red} - 16}} \right) - \left( {{{\left( {y + 2} \right)}^2} \,{\color{ProcessBlue} - 4}} \right) + 24 & = 0\\ 4{\left( {x - 4} \right)^2} - {\left( {y + 2} \right)^2} + 24 \,{\color{Red} - 64} \,{\color{ProcessBlue} + 4} & = 0\\ 4{\left( {x - 4} \right)^2} - {\left( {y + 2} \right)^2} - 36 & = 0\end{align*}\]Konstantalarni yuqoriga qo'shayotganda, konstantalarni qo'shishdan oldin 4 ni \(x\) shartlariga va -1 ni \(y\) shartlariga ko'paytirishni unutmang.
4-qadam
Ishni tugatish uchun avval 36 ni tenglamaning boshqa tomoniga o'tkazamiz.
\[4{\left( {x - 4} \right)^2} - {\left( {y + 2} \right)^2} = 36\]Buni standart shaklga keltirish uchun bizga tenglamaning o'ng tomonidagi bittasi kerak. Buni olish uchun hamma narsani 36 ga bo'lish kifoya va biz \(x\) atamasi bo'yicha biroz soddalashtirish ishlarini bajaramiz.
\[\frac{{4{{\left( {x - 4} \right)}^2}}}{{36}} - \frac{{{{\left( {y + 2} \right)}^2}}}{{36}} = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{{\left( {x - 4} \right)}^2}}}{9} - \frac{{{{\left( {y + 2} \right)}^2}}}{{36}} = 1}}\]