Eisimpleir de Cheistean Deasbaid Geoimeatraidh Anailiseach
Pendahuuan
’S e meur de mhatamataig a th’ ann an geoimeatraidh anailiseach a bhios a’ cothlamadh ailseabra agus geoimeatraidh gus fuasgladh fhaighinn air duilgheadasan anns a bheil àite agus cumadh. ’S e inneal cumhachdach a th’ ann a leigeas leinn duilgheadasan geoimeatrach a sgrùdadh le bhith a’ cleachdadh cho-aontaran agus co-chomharran. Bruidhnidh an t-artaigil seo air grunn eisimpleirean de dhuilgheadasan farsaingeachd geoimeatraidh anailiseach agus bruidhnidh e orra gu mionaideach gus tuigse nas doimhne a dhèanamh nas fhasa.
Eisimpleir Ceist 1: Co-aontar Loidhne
Ceist:
Ma tha dà phuing A(1, 2) agus B(3, 7) air an toirt seachad, obraich a-mach co-aontar na loidhne a tha a’ dol tron dà phuing seo.
Deasbad:
Gus co-aontar loidhne a lorg a tha a’ dol tro dhà phuing, is urrainn dhuinn foirmle a’ chlaonaidh (leòid) m a chleachdadh:
[m = \frac{y_2 – y_1}{x_2 – x_1}]
Le puing A(x1, y1) = (1, 2) agus puing B(x2, y2) = (3, 7):
[m = \frac{7 – 2}{3 – 1} = \frac{5}{2} \]
An ath rud, cleachdaidh sinn am foirmle airson co-aontar na loidhne:
[y – y_1 = m(x – x_1)]
Ionadachadh aon phuinge, mar eisimpleir puing A(1, 2):
[y – 2 = \frac{5}{2}(x – 1) \]
Tionndaidh an cruth seo gu co-aontar soilleir airson y:
[y – 2 = \frac{5}{2}x – \frac{5}{2} \]
[y = \frac{5}{2}x – \frac{5}{2} + 2 \]
[y = \frac{5}{2}x – \frac{1}{2} \]
Mar sin, is e co-aontar na loidhne:
[y = \frac{5}{2}x – \frac{1}{2} \]
Eisimpleir Ceist 2: Cearcall
Ceist:
Obraich a-mach co-aontar cearcaill aig a bheil meadhan aig puing C(-2, 3) agus le radius de 4.
Deasbad:
Is e co-aontar coitcheann cearcaill le meadhan aig (h, k) agus radius r:
[(x – h)^2 + (y – k)^2 = r^2]
A rèir na ceiste, meadhan a’ chearcaill (h, k) = (-2, 3) agus an radius r = 4. Mar sin,
\[(x + 2)^2 + (y – 3)^2 = 4^2 \]
\[(x + 2)^2 + (y – 3)^2 = 16 \]
Mar sin, is e co-aontar a’ chearcaill:
\[(x + 2)^2 + (y – 3)^2 = 16 \]
Eisimpleir Ceist 3: Parabola
Ceist:
Obraich a-mach co-aontar parabola dìreach leis a’ mhullach aig (1, -2) agus am fòcas aig (1, 0).
Deasbad:
Airson parabola dìreach le bàrr (h, k), is e seo an co-aontar coitcheann:
[(x – h)^2 = 4p(y – k)]
Ma tha an vertex (h, k) = (1, -2), feumaidh sinn luach p a lorg. Is e (h, k + p) fòcas a’ pharabola, agus bhon duilgheadas is e (1, 0) am fòcas:
[k + p = 0 – (-2) = 2]
Mar sin:
\[p = 2 \]
Mar sin, bidh an co-aontar coitcheann mar a leanas:
[(x – 1)^2 = 4 ⋅2 (y + 2)]
[(x – 1)^2 = 8(y + 2)]
Mar sin, is e co-aontar a’ pharabola:
[(x – 1)^2 = 8(y + 2)]
Eisimpleir Ceist 4: Ellips
Ceist:
Ma tha eileips ann le meadhan aig puing (0, 0), fad a’ phrìomh ais 10 agus an ais bheag 6. Obraich a-mach co-aontar an eileips.
Deasbad:
Is e meadhan an eileips (h, k) (0, 0), fad na prìomh ais 2a = 10 gus am bi a = 5, agus fad na h-ais bhig 2b = 6 gus am bi b = 3. Is e seo an co-aontar coitcheann airson eileips le meadhan aig (0, 0):
[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Cuir luachan a agus b an àite:
[ \frac{x^2}{5^2} + \frac{y^2}{3^2} = 1 \]
[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]
Mar sin, is e co-aontar an ellipse:
[ \frac{x^2}{25} + \frac{y^2}{9} = 1 \]
Eisimpleir Ceist 5: Hipearbola
Ceist:
Ma tha hipearbola aig a bheil meadhan aig (1, -3), tha fad an ais thrasnach 8 agus fad an ais cho-cheangailte 6. Obraich a-mach co-aontar an hipearbola.
Deasbad:
Airson hipearbola le meadhan (h, k) agus ais-tharraingeach chòmhnard, is e seo an co-aontar coitcheann:
[ \frac{(x – h)^2}{a^2} – \frac{(y – k)^2}{b^2} = 1 \]
Is e meadhan an hipearbola (h, k) (1, -3), is e fad an ais thrasnach 2a = 8 agus mar sin tha a = 4, agus is e fad an ais cho-cheangailte 2b = 6 agus mar sin tha b = 3. Mar sin, is e co-aontar an hipearbola:
[ \frac{(x – 1)^2}{4^2} – \frac{(y + 3)^2}{3^2} = 1 \]
[ \frac{(x – 1)^2}{16} – \frac{(y + 3)^2}{9} = 1 \]
Mar sin, is e co-aontar an hipearbola:
[ \frac{(x – 1)^2}{16} – \frac{(y + 3)^2}{9} = 1 \]
Co-dhùnadh
’S e dòigh chumhachdach a th’ ann an geoimeatraidh anailiseach airson cumaidhean agus structaran geoimeatrach a sgrùdadh le bhith a’ cleachdadh co-aontaran ailseabrach. Le bhith a’ tuigsinn bun-bheachdan co-aontaran loidhnichean, chearcaill, pharabolan, eileipsean, agus hipearbolan, is urrainn dhuinn measgachadh de dhuilgheadasan geoimeatraidh fhuasgladh gu furasta. Tha an t-artaigil seo a’ toirt seachad eisimpleirean agus deasbadan mu dhuilgheadasan cudromach ann an geoimeatraidh anailiseach gus do chuideachadh le bhith a’ doimhneachadh do thuigse. Faodaidh duilgheadasan cleachdaidh a bharrachd do chuideachadh le bhith a’ neartachadh agus a’ leudachadh do thuigse air an stuth seo.
