Isibonelo Sombuzo Wengxoxo Nge-Equation Yendilinga
I-equation yendilinga iyisihloko esibalulekile ku-analytical geometry. Ukuqonda kahle i-equation yendilinga kuwusizo kakhulu, hhayi kuphela kwizibalo kodwa futhi nasezinhlelweni ezahlukene zobunjiniyela nezesayensi. Kulesi sihloko, sizoxoxa ngezibonelo eziningana ze-equation yendilinga kanye nezixazululo zazo. Umgomo ukunikeza umbono ocacile nophelele wendlela yokuxazulula izinkinga ezihilela i-equation yendilinga.
Isibalo Esijwayelekile Sendilinga
I-equation evame kakhulu yesiyingi kuma-coordinates e-Cartesian yile:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Di mana:
– \( (a, b) \) yizixhumanisi zesikhungo sendilinga.
– \( r \) irediyasi yesiyingi.
Uma isikhungo sendilinga sisephuzwini \( (0, 0) \), isibalo sendilinga sizoba:
\[ x^2 + y^2 = r^2 \]
Manje, ake sixoxe ngemibuzo ethile eyisibonelo kanye nezixazululo zayo.
Isibonelo Umbuzo 1
Umbuzo: Thola i-equation yendilinga enendawo ephakathi nendawo (3, -2) futhi enobubanzi obungu-5.
Isixazululo:
Sebenzisa ifomula ejwayelekile yesibalo sendilinga:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Faka amanani esikhundleni sawo \( a = 3 \), \( b = -2 \), kanye \( r = 5 \):
\[ (x – 3)^2 + (y + 2)^2 = 5^2 \]
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Ngakho-ke, isibalo sendilinga sithi:
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Isibonelo Umbuzo 2
Umbuzo: Thola i-equation yendilinga enendawo ephakathi kwayo esekuqaleni (0, 0) futhi inobubanzi obungu-7.
Isixazululo:
Njengoba isikhungo sendilinga sisekuqaleni, singasebenzisa i-equation elula:
\[ x^2 + y^2 = r^2 \]
Faka inani esikhundleni \(r = 7 \):
\[ x^2 + y^2 = 7^2 \]
\[ x^2 + y^2 = 49 \]
Ngakho-ke, isibalo sendilinga sithi:
\[ x^2 + y^2 = 49 \]
Isibonelo Umbuzo 3
Umbuzo: Thola isibalo sendilinga esiphakathi kwayo sisephuzwini (4, -5) futhi sithinta i-axis ka-Y.
Isixazululo:
I-tanjent yesiyingi ku-Y-axis isho ukuthi ibanga elisuka enkabeni yesiyingi liye ku-Y-axis lilingana ne-radius yalo. Leli banga liyinani eliphelele le-X-coordinate yenkaba yesiyingi. Ngakho-ke, i-radius ingu-4.
Sebenzisa ifomula ejwayelekile yesibalo sendilinga:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Faka amanani esikhundleni sawo \( a = 4 \), \( b = -5 \), kanye \( r = 4 \):
\[ (x – 4)^2 + (y + 5)^2 = 4^2 \]
\[ (x – 4)^2 + (y + 5)^2 = 16 \]
Ngakho-ke, isibalo sendilinga sithi:
\[ (x – 4)^2 + (y + 5)^2 = 16 \]
Isibonelo Umbuzo 4
Umbuzo: Indilinga ine-equation \( x^2 + y^2 – 6x + 4y – 12 = 0 \). Thola isikhungo kanye nerediyasi yendilinga.
Isixazululo:
Ukuze sixazulule lesi sibalo, sidinga ukusiguqula sibe yifomu ejwayelekile \( (x – a)^2 + (y – b)^2 = r^2 \). Izinyathelo zokusiqedela yilezi ezilandelayo:
1. Ukuqoqa nokuxazulula izikwele eziphelele:
Isilinganiso sokuqala sithi:
\[ x^2 + y^2 – 6x + 4y – 12 = 0 \]
Iqembu \( x \) kanye \( y \):
\[ (x^2 – 6x) + (y^2 + 4y) = 12 \]
2. Xazulula isikwele esiphelele:
Ukuze \( x^2 – 6x \):
\[ x^2 – 6x + 9 \]
Ngoba \( y^2 + 4y \):
\[ y^2 + 4y + 4 \]
Engeza u-9 no-4 ezinhlangothini zombili ze-equation:
\[ (x^2 – 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 \]
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Ngakho-ke, isibalo sendilinga ngesimo esijwayelekile yilesi:
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Kusukela lapha, singabona ukuthi isikhungo sendilinga siyi-\( (3, -2) \) kanti irediyasi iyi-\( r = \sqrt{25} = 5 \).
Isibonelo Umbuzo 5
Umbuzo: Thola isibalo sendilinga edlula emaphuzwini (2, 3) kanye no-(4, 5), futhi isikhungo saso sisemgqeni u-x = 3.
Isixazululo:
Kusukela embuzweni, siyazi ukuthi isikhungo sendilinga yi-(3, b). Indilinga idlula namaphuzu amabili aziwayo. Njengoba indilinga idlula ku-(2, 3), ibanga ukusuka enkabeni kuya kuleli phuzu liyi-radius.
Isibalo sendilinga sithi:
\[ (x – 3)^2 + (y – b)^2 = r^2 \]
Iphuzu lokufaka esikhundleni (2, 3):
\[ (2 – 3)^2 + (3 – b)^2 = r^2 \]
\[ 1 + (3 – b)^2 = r^2 \]
\[ (3 – b)^2 = r^2 – 1 \]
Iphuzu lokufaka esikhundleni (4, 5):
\[ (4 – 3)^2 + (5 – b)^2 = r^2 \]
\[ 1 + (5 – b)^2 = r^2 \]
\[ (5 – b)^2 = r^2 – 1 \]
Kusukela kulezi zibalo ezimbili, siyazi ukuthi (3 – b)^2 = (5 – b)^2. Ngakho-ke:
\[ 3 – b = \pm(5 – b) \]
Uma \( 3 – b = 5 – b \), umphumela awukwazi ukuba yiqiniso. Ngakho-ke:
\[ 3 – b = -(5 – b) \]
\[ b = 4 \]
Nge-b = 4, isibalo sendilinga sithi:
\[ (x – 3)^2 + (y – 4)^2 = 2 \]
Noma kunjalo, singabala i-radius r kusukela ebangeni eliphakathi nendawo kanye nephuzu (2, 3) = \(\sqrt{(2 – 3)^2 + (3 – 4)^2} \) = \(\sqrt{1+1}\) = \(\sqrt {2}\)
Isibalo sendilinga sithi:
\[ (x – 3)^2 + (y – 4)^2 = 2 \]
Isiphetho
Ukuqonda i-equation yendilinga kungenza kube lula ukuxazulula izinkinga eziningi zezibalo. Esimweni ngasinye, ukuhlonza isikhungo kanye ne-radius kubalulekile. Ngethemba ukuthi lezi zinkinga zezibonelo kanye nezincazelo zazo zinikeza ukucaca futhi zikusize ufunde i-equation yendilinga. Ukuzijwayeza kwenza kube kuhle kakhulu ezibalweni, ngakho ungangabazi ukuzama izinkinga ezahlukahlukene ukuze uthuthukise amakhono akho.
