Mehlala ea Lipotso tse Buisanang ka Likarolo tsa Hyperbolic Conic
Pendahuluan
Lipalong tsa lipalo, karolo ea conic, eo hangata e bitsoang karolo ea conic, ke mothinya o fumanoang moo khoune le sefofane li kopanang teng. Ho na le mefuta e mene e meholo ea likarolo tsa conic: li-circles, li-ellipses, li-parabola, le li-hyperbola. Sehloohong sena, re tla tsepamisa maikutlo ho hyperbola, mofuta oa karolo ea conic e nang le lits'ebetso tse ngata masimong a kang bolepi ba linaleli, fisiks le boenjiniere. Sehlooho sena se tla hlahisa mehlala ea mathata le puisano ea bona ka taba ena, ka sepheo sa ho thusa babali ho utloisisa mohopolo le mokhoa oa ho rarolla mathata a amanang le hyperbolas.
Tlhaloso le Matlotlo a Hyperbole
Pele re kena lipotsong tsa mohlala, a re qaleng ka ho buisana ka likhopolo tse ling tsa motheo mabapi le hyperbola.
Hyperbola ke sebaka sa lintlha tse sefofaneng hoo phapang ea sebaka sa ntlha ka 'ngoe ho tloha lintlheng tse peli tse tsitsitseng (tse bitsoang foci) e sa fetoheng.
Tekanyo e akaretsang ea hyperbola ka mokhoa o tloaelehileng ke:
\[ \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 \]
kapa
\[ \frac{y^2}{b^2} – \frac{x^2}{a^2} = 1 \]
Moo:
– \(a\) ke sebaka se tlohang bohareng ba hyperbola ho ya tlhorong ya yona (vertex).
– \(b\) ke sebaka se amanang le sebaka se tlohang bohareng ho ya ntlheng e haufi ho asymptote ya hyperbola.
Bakeng sa hyperbola e tšelang ka ho otloloha, sebopeho se akaretsang se sebelisitsoeng ke:
\[ \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 \]
Ho sa le jwalo, bakeng sa di-hyperbola tse tshelang ka ho otloloha:
\[ \frac{y^2}{b^2} – \frac{x^2}{a^2} = 1 \]
Lipotso tsa Mehlala le Puisano
Potso ea 1:
Ha ho fanoe ka equation ea hyperbola \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \). Fumana:
1. Setsi sa hyperbola.
2. Bolelele ba mothapo o ka sehloohong le mothapo o mong.
3. Ntlha e shebaneng le ntlha.
4. Tekanyo ea Asymptote.
5. Thala hyperbola.
Puisano:
1. Setsi sa Hyperbola:
Kaha sebopeho sa equation e ka hodimo ke sa maemo a hodimo mme ha ho na mantswe \((x – h)\) kapa \((y – k)\), setsi sa hyperbola ena se ntlheng ya (0,0).
2. Bolelele ba Axis e Kholo le Axis ea Bobeli:
Ho tsoa ho equation \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \), ho tsebahala hore:
\[
a^2 = 16 \Motsu o ka letsohong le letona a = 4
\]
\[
b^2 = 9 \Motsu o ka letsohong le letona b = 3
\]
Bolelele ba mothapo o moholo ke \(2a = 2 \makgetlo a 4 = 8\).
Bolelele ba axis ya bobedi ke \(2b = 2 \makgetlo a 3 = 6\).
3. Ntlha ea bohlokoa:
Ho fumana ntlha ea bohlokoa, re sebelisa kamano ena:
\[
c^2 = a^2 + b^2
\]
\[
c^2 = 16 + 9 = 25 \Motsu o letona c = \sqrt{25} = 5
\]
Kaha hyperbola ena e rapame, li-foci li lintlheng tsa \((\pm c, 0)\), e leng \((5, 0)\) le \((-5, 0)\).
4. Tekanyo ea Asymptote:
Asymptote ke mola o otlolohileng o atamelang hyperbola. Bakeng sa equation ena e tloaelehileng, asymptote e ka fumanoa ka:
\[
y = \pm \frac{b}{a}x \Rightarrow y = \pm \frac{3}{4}x
\]
Kahoo, di-equation tsa asymptote ke \( y = \frac{3}{4}x \) le \( y = -\frac{3}{4}x \).
5. Setšoantšo sa Hyperbola:
Ho hlalosa hyperbola, re hloka:
– E tšoaea bohareng ho (0,0).
– Hlophisa litlhōrō lintlheng (4,0) le (-4,0).
– Thala di-asymptotes ka mela ya y = (3/4)x le y = -(3/4)x e fetang bohareng.
– Tshwaya dintlha tsa tlhokomelo ho (5,0) le (-5,0).
Potso ea 2:
Fumana equation ea hyperbola e nang le axis e kholo ea bolelele ba liyuniti tse 10, axis ea bobeli ea bolelele ba liyuniti tse 8, 'me e bohareng ba tšimoloho.
Puisano:
Ho tsoa potsong ho tsebahala hore bolelele ba axis e kholo (2a) ke diyuniti tse 10, ebe:
\[ 2a = 10 \Motsu a = 5 \]
Bolelele ba axis ea bobeli (2b) ke liyuniti tse 8, kahoo:
\[ 2b = 8 \Motsu o ka letsohong le letona b = 4 \]
Ka setsi qalong (0,0), re ka ngola equation e tloaelehileng ea hyperbola ka tsela e latelang:
\[ \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 \]
Kamora ho nkela boleng ba a le b sebaka:
\[ \frac{x^2}{25} – \frac{y^2}{16} = 1 \]
Kahoo, equation ea hyperbola e botsoang ke:
\[ \frac{x^2}{25} – \frac{y^2}{16} = 1 \]
Potso ea 3:
Ha ho fanoe ka hyperbola e otlolohileng e nang le equation \(\frac{y^2}{36} – \frac{x^2}{16} = 1 \). Fumana sebaka se pakeng tsa li-foci tsa eona tse peli.
Puisano:
Bakeng sa equation ea hyperbola \(\frac{y^2}{36} – \frac{x^2}{16} = 1\), re khetholla hore:
\[ a^2 = 36 \Motsu o ka letsohong le letona a = 6 \]
\[ b^2 = 16 \Motsu o ka letsohong le letona b = 4 \]
Ho fumana sebaka se pakeng tsa li-foci tse peli, re sebelisa kamano:
\[ c^2 = a^2 + b^2 \]
\[ c^2 = 36 + 16 = 52 \Motsu o letona c = \sqrt{52} = 2\sqrt{13} \]
Sebaka se pakeng tsa li-foci tse peli tsa hyperbola se baloa ka makhetlo a mabeli ho feta sebaka sa foci ho tloha bohareng:
\[ 2c = 2 \makgetlo a 2\sqrt{13} = 4\sqrt{13} \]
Kahoo, sebaka se pakeng tsa tse peli tse shebaneng ke diyuniti tse \(4\sqrt{13}\).
Qetello
Sehloohong sena, re buisane ka mehlala e 'maloa ea mathata mabapi le li-hyperbola, ho kenyeletsoa ho khetholla setsi, bolelele ba li-axes tse kholo le tse nyane, foci, li-equation tsa li-asymptotes, le grafing ea li-hyperbolas. Ho utloisisa mokhoa oa ho rarolla mathata ana ho bohlokoa, haholo-holo bakeng sa baithuti ba ithutang jiometri ea tlhahlobo kapa lipalo tse tsoetseng pele.
Hyperbola ha se khopolo-taba feela, empa hape e na le lits'ebetso tse pharaletseng mafapheng a mang a saense joalo ka astrophysics, radar le GPS. Ka hona, ho ithuta hyperbola ha se feela ho rarolla mathata a lipalo, empa hape le ho utloisisa kamoo likhopolo tsena tsa lipalo li ka sebelisoang kateng bophelong ba sebele.
