Imibuzo Eyisibonelo Exoxa Ngemibuthano Nezingqinamba
Izindilinga ziyisihloko esibalulekile ku-matrix geometry, lapho kuboniswa khona imiqondo ejulile mayelana nebanga, ama-engeli, kanye nesimo. Omunye umqondo ovame ukuxoxwa ngawo kulesi sihloko umugqa we-tangent oya embuthanweni. Kulesi sihloko, sizoxoxa ngezinkinga eziningana zezibonelo ezihilela izindilinga nama-tangent.
Ukuqonda Okuyisisekelo Kwemibuthano Nezingqinamba
Umbuthano
Indilinga iyisimo sejiyometri esakhiwe yiqoqo lamaphuzu onke endizeni ayibanga eliqondile ukusuka endaweni ethile ebizwa ngokuthi isikhungo sendilinga. Leli banga eliqondile libizwa ngokuthi i-radius yendilinga.
I-Tangent
I-tangent embuthanweni umugqa othinta indilinga endaweni eyodwa ngqo. Leli phuzu libizwa ngokuthi iphuzu le-tangency. Ama-tangent anezakhiwo eziningana ezibalulekile, okuhlanganisa:
– Umugqa oqondile uhlala uqonde ngqo erediyasini yendilinga endaweni lapho uqondile khona.
– Ubude be-tangent kusukela endaweni engaphandle kwendilinga kuya endilinga buyafana uma kudwetshwa ama-tangent amabili kuleyo ndawo.
Imibuzo Nezingxoxo Eziyisibonelo
Ngezansi sizokwethula imibuzo eminingana eyisibonelo exoxa ngomqondo wezindilinga nama-tangent ngokuningiliziwe.
Isibonelo Umbuzo 1: Ukuthola Ubude Bomugqa Oqondile
Umbuzo:
Uma unikezwe indilinga enesikhungo \(O\) kanye nerediyasi \(r = 6 \, \text{cm}\). Kusukela endaweni \(P\) ngaphandle kwendilinga engama-10 cm ukusuka enkabeni yendilinga, kudonswa ama-tangent amabili \(PA\) kanye \(PB\) endilinga. Bala ubude be-tangent \(PA\).
Ingxoxo:
Kule nkinga, singasebenzisa i-Pythagorean theorem. Dweba unxantathu \(\unxantathu OAP\):
– \(OP = 10 \, \text{cm}\) (ibanga ukusuka endaweni engaphandle kuya enkabeni yendilinga)
– \(OA = 6 \, \text{cm}\) (irediyasi yendilinga)
– \(PA\) umugqa oqondile okumele utholakale
\[
OP^2 = OA^2 + PA^2
\]
\[
10^2 = 6^2 + PA^2
\]
\[
100 = 36 + PA^2
\]
\[
I-PA^2 = 64
\]
\[
I-PA = \sqrt{64} = 8 \, \text{cm}
\]
Ngakho-ke, ubude bomugqa we-tangent \(PA\) buyi-8 cm.
Isibonelo Umbuzo 2: Ukuthola Iphuzu Lokuthambekela
Umbuzo:
Unikezwe indilinga ene-equation \((x – 3)^2 + (y – 4)^2 = 25\) kanye nomugqa \(y = 2x + 1\). Thola iphuzu lokuhlangana phakathi kwendilinga nomugqa.
Ingxoxo:
Okokuqala, sithola isikhungo kanye nerediyasi yendilinga:
– Isikhungo \(O(3, 4)\)
– Irediyasi \(r = \sqrt{25} = 5\)
Ukuze sithole iphuzu le-tangency, ake sithi iphuzu le-tangency yi-\(T(x_1, y_1)\) nayo elele emgqeni \(y = 2x + 1\). Bese:
\[
y_1 = 2x_1 + 1
\]
\(T(x_1, y_1)\) kumele futhi ihlangabezane nesibalo sendilinga:
\[
(x_1 – 3)^2 + (y_1 – 4)^2 = 25
\]
Faka u-\(y_1 = 2x_1 + 1\) esilinganisweni sesiyingi:
\[
(x_1 – 3)^2 + ((2x_1 + 1) – 4)^2 = 25
\]
\[
(x_1 – 3)^2 + (2x_1 – 3)^2 = 25
\]
Sidinga ukubala izikwele ezimbili.
\[
(x_1 – 3)^2 = x_1^2 – 6x_1 + 9
\]
\[
(2x_1 – 3)^2 = 4x_1^2 – 12x_1 + 9
\]
Hlanganisa imiphumela yomibili:
\[
x_1^2 – 6x_1 + 9 + 4x_1^2 – 12x_1 + 9 = 25
\]
\[
5x_1^2 – 18x_1 + 18 = 25
\]
Susa u-25 kuzo zombili izinhlangothi:
\[
5x_1^2 – 18x_1 – 7 = 0
\]
Xazulula i-quadratic equation:
\[
x_1 = \frac{18 \pm \sqrt{18^2 + 4 \times 5 \times 7}}{2 \times 5}
\]
\[
x_1 = \frac{18 \pm \sqrt{324 + 140}}{10}
\]
\[
x_1 = \frac{18 \pm \sqrt{464}}{10}
\]
\[
x_1 = \frac{18 \pm 2\sqrt{116}}{10}
\]
\[
x_1 = \frac{18 \pm 2\sqrt{4 \times 29}}{10}
\]
\[
x_1 = \frac{18 \pm 4\sqrt{29}}{10}
\]
\[
x_1 = 1.8 \pm 0.4\sqrt{29}
\]
Bala inani le-\(y_1\):
Okwanelisa u-y = 2x + 1:
– Uma \(x_1 = 1.8 + 0.4\sqrt{29}\), khona-ke \(y_1 = 2(1.8 + 0.4\sqrt{29}) + 1\)
– Uma \(x_1 = 1.8 – 0.4\sqrt{29}\), khona-ke \(y_1 = 2(1.8 – 0.4\sqrt{29}) + 1\)
Ukuhlolwa:
Ngakho-ke sithola amaphuzu amabili okuhlangana kwesibalo sendilinga nalowo mugqa.
Isibonelo Umbuzo 3: Ukunquma Isibalo Somugqa Oqondile
Umbuzo:
Unikezwe indilinga ene-equation \((x – 2)^2 + (y – 3)^2 = 20\). Thola i-equation yomugqa we-tangent endilinga edlula ephuzwini \((6, 7)\).
Ingxoxo:
I-tangent kumbuthano onesikhungo \((h, k)\) kanye nerediyasi \(r\) kusuka endaweni yangaphandle eyaziwayo ingatholakala nge-equation:
Umugqa we-tangent udlula endaweni yangaphandle \((x_1, y_1)\):
\[
(x – 2)(x_1 – 2) + (y – 3)(y_1 – 3) = 20
\]
Faka esikhundleni sephuzu elingaphandle \((6, 7)\):
\[
(x – 2)(6 – 2) + (y – 3)(7 – 3) = 20
\]
\[
4(x – 2) + 4(y – 3) = 20
\]
\[
4(x – 2 + y – 3) = 20
\]
\[
4x + 2y -20 = 20
\]
\[
4x + 4y -20 = 20
\]
\[
x + y = 5
\]
Isibalo somugqa we-tangent sithi:
\[
x + y = 9
\]
Ngakho-ke, ukwehluka kwesibalo somugqa odlula endaweni yomugqa wesiyingi kukhulu kakhulu futhi kungashintsha kuye ngomphumela noma ukumelwa okubonakalayo.
Isiphetho
Ingxoxo ngezindilinga nama-tangent ihlanganisa izici eziningana eziyisisekelo zezibalo, kusukela ekusebenziseni amafomula ayisisekelo njenge-Pythagorean theorem kuya ekuxazululeni ama-quadratic equations. Ngalezi zibonelo, singathuthukisa ukuqonda okungcono kokuthi singayisebenzisa kanjani le mibono ezimweni eziyinkimbinkimbi. Ngethemba ukuthi lesi sihloko sisize ekuhlinzekeni isithombe esicacile sendlela yokusondela nokuxazulula izinkinga ezihilela izindilinga nama-tangent.
