Su'aalo Tusaale ah oo Ka Hadlaya Isle'egta Xariiqda Tangent-ka ee Goobo
Pendahuluan
Isle'egta tangent-ka ee goobada waa mowduuc muhiim ah oo ku jira joomatari falanqayn ah. Fahmidda sida loo go'aamiyo isle'egta tangent-ka ee goobada waxay kaa caawin kartaa xallinta dhibaatooyin xisaabeed oo kala duwan oo heer dhexe ilaa heer sare ah. Maqaalkani wuxuu ka hadli doonaa masalooyin kala duwan iyo habab lagu go'aaminayo isle'egta tangent-ka ee goobada.
Qeexidda iyo Aragtida Aasaasiga ah
Goobo ku yaal diyaaradda isku-dhafka ah waxaa badanaa lagu matali karaa isla'eg labajibbaaran:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
halkaas oo \((a, b)\) ay tahay barta dhexe ee goobada iyo \(r\) ay tahay gacanka goobada.
Taangent-ka goobada laga soo bilaabo barta dibadda waa xariiq taabanaysa goobada hal dhibic. Haddii aan u qaadanno in xariiqdu ay leedahay isla'eg:
\[y = mx + c \]
markaas shuruudda ah in xariiqda \(y = mx + c\) ay la jaanqaadi doonto goobada waxaa lagu muujin karaa qaabkan soo socda:
\[ \sqrt{(a + bm)^2 – r^2} = |c| \]
Halka \(m\) uu yahay jahada xariiqda tangent iyo \(c\) uu yahay joogto.
Qaacidada Isle'egta Khadka Tangent
Taangent-ka goobada leh bartamaha \((0,0)\) iyo radius \(r\), isle'egtiisa barta \((x_1, y_1)\) ee goobada waa:
\[ x_1x + y_1y = r^2 \]
Laakiin haddii bartamaha goobada ay ku taal barta \((a, b)\), markaas isle'egtu waa:
\[ (x_1 – a)(x – a) + (y_1 – b)(y – b) = r^2 \]
Su'aalo iyo Doodo Tusaale ah
Su'aal 1aad
Goobo leh xarun ku taal \((3, 4)\) iyo gacan 5 cutub. Go'aami isle'egta xariiqda tangent laga bilaabo barta \((6, 8)\).
Dood:
Dhibaatadan, waxaa naloo siiyay goobaabin leh xarun \((3, 4)\), gacan 5 ah, waana inaan helnaa isle'egta xariiqda tangent laga bilaabo barta \((6, 8)\). Waa kuwan tallaabooyinka lagu xallinayo:
1. Hubi in barta \((6, 8)\) aysan ku jirin goobada: \\
\[
\sqrt{(6-3)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5
\]
Markaa bartu waxay ku taal goobada sidaa darteed waxaa loo isticmaali karaa in lagu helo xariiqda tuujinta.
2. Isticmaalka qaacidada:
\[
(x_1 – a)(x – a) + (y_1 – b)(y – b) = r^2
\]
\[
(6 – 3) (x – 3) + (8 – 4) (y – 4) = 5^2
\]
3. Fududee isla'egta:
\[
3(x – 3) + 4(y – 4) = 25
\]
4. Horumari:
\[
3x – 9 + 4y – 16 = 25
\]
\[
3x + 4y - 25 = 50
\]
Isle'egta xariiqda tangent-ka ee la helay waa:
\[
3x + 4y = 50
\]
Su'aal 2aad
Goobo leh isla'eg \[x^2 + y^2 = 16\]. Go'aami isle'egta xariiqda tangent laga bilaabo barta \((4, 0)\).
Dood:
Goobo leh xarun ku taal \((0, 0)\) iyo gacan 4 cutub. Barta dibadda ee la bixiyay waa \((4, 0)\).
1. Isticmaalka qaacidada goobaabin leh bartamaha \((0, 0)\):
\[
x_1x + y_1y = r^2
\]
2. Beddelka qiimaha:
\[
4x + 0 \cdot y = 4^2
\]
3. Fududee:
\[
4x = 16
\]
\[
x = 4
\]
Taas macnaheedu waa, isla'egta tangent ee aan haysanno waa:
\[
x = 4
\]
Su'aal 3aad
Samee xariiq la jaanqaadaysa goobada \((x+2)^2 + (y-3)^2 = 9\) barta \(-1, 5)\).
Dood:
Goobo leh xarun ku taal \((-2, 3)\) iyo gacanka 3 cutub. Barta isku-dhafka waa \((-1, 5)\).
1. Hubi in barta ay ku taal goobada:
\[
((-1+2)^2 + (5-3)^2) = 1 + 4 = 5 \neq 9 \rightarrow \text{kuma jiraan goobada}
\]
Sidaas darteed, waxaa jiri kara qalad ku jira gelinta su'aasha ama dhibicda. Haddii \((-2, 6)\) loo bixiyo barta isku-dhafka ah:
2. Fududee:
\[
(x_1-a)(xa) + (y_1-b)(yb) = r^2
\]
Beddelka:
\[
(-2-(-2)) (x+2) + (6-3) (y-3) = 9\]
Natiijo:
\[
0 + 3(s-3) = 3^2
\]
\[
3 sano-9 = 9
\]
\[
y = 6
Waxaa soo baxday in qodobka & pers.:
y = 6 waa hubaal sax.
Isku-darka hubinta gunaanadka, xifdi, 2 siyaabood oo loo tiriyo.
sax ah/been..Taas oo lagu daray soo koobid gar. td 2; qaab sax ah oo leh (shabakad). plis wareega dhammaan daboolida btns.
Intaas oo dhan, waad ku mahadsan tahay waxaana rajeynayaa inay tani faa'iido u leedahay asxaabtayda.
Intaa waxaa dheer, baro fikradda xoojinta fahamka.
Kuwa aan hore u fahmin, haddii Alle idmo, waxay yeelan doonaan tallaabooyin qalliin oo faahfaahsan oo faahfaahsan.
\[
.facts(n+x)=qa.\color{prevents}-fu.
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