Ngā Tauira Pātai me te Kōrero mō ngā Pānga ki ngā Wāhanga Kōnika
Pendahuluan
Ko te wāhanga kōniko he kōpiko e puta mai ana i te whakawhitiwhitinga o tētahi paparangi me tētahi kōniko rua-polar. Kei roto i ēnei kōpiko ngā porowhita, ngā porowhita, ngā parabola, me ngā hyperbola. Ko tētahi kaupapa nui hei mārama ki ngā wāhanga kōniko ko te rārangi pātata. Ko te pātata ki tētahi wāhanga kōniko he rārangi e pā ana ki te kōpiko kōniko i te kotahi pūwāhi anake. Ka matapakihia e tēnei tuhinga ētahi tauira raruraru me te matapakinga o ngā pātata ki ngā wāhanga kōniko.
Pānga ki te Porowhita
He wāhanga kōniko te porowhita, he āhua māmā noa iho, ā, he ōrite tino tika. Me tīmata tātou me tētahi tauira raruraru mō ngā pātapa ki tētahi porowhita.
Tauira Pātai 1
Hoatu he porowhita me te whārite \( (x – 2)^2 + (y + 3)^2 = 25 \). Whakatauhia te whārite o te rārangi pātapa i te pūwāhi \((5, -3)\) i runga i te porowhita.
Kōrero
Ko te whārite whānui o tētahi porowhita ko \( (x – h)^2 + (y – k)^2 = r^2 \), ko \( (h, k) \) te pokapū o te porowhita, ā, ko \( r \) te pūtoro. I roto i tēnei raruraru, ko te pokapū o te porowhita \((h, k)\) ko \((2, -3)\) ā, ko te pūtoro \( r = \sqrt{25} = 5 \).
Ka kitea te rārangi pātapa i te pūwāhi \((x_1, y_1)\) i runga i te porowhita mā te whakamahi i te tātai e whai ake nei:
\[ (x – h)(x_1 – h) + (y – k)(y_1 – k) = r^2 \]
Whakauruhia ngā uara e mōhiotia ana:
\[ (x – 2)(5 – 2) + (y + 3)(-3 + 3) = 25 \]
\[ (x – 2)(3) + (y + 3)(0) = 25 \]
\[ 3(x – 2) = 25 \]
\[ 3x – 6 = 25 \]
\[ 3x = 31 \]
\[ x = \frac{31}{3} \]
Ko te whārite o te rārangi pātata ko \(x = \frac{31}{3}\), engari he hapa kei roto i tēnei tikanga nā te mea ko te pūwāhi \((5, -3)\) he pūwāhi mārama kei runga i te porowhita. Nō reira, ka whakamahia e mātou te tikanga tuku iho mā te whakakapi i te pari o te rārangi pātata i tēnei pūwāhi motuhake:
Ko te pūwāhi pātata ko, \((5, -3)\), kātahi, ko te pikinga (m) o te rārangi pūtoro ko \(m = \frac{-3 – (-3)}{5 – 2}=0\), ina kore e tautuhia te pikinga o te rārangi pātata mō te pātata poutū o mua.
Rārangi Pānga ki te Porowhita
He wāhanga kōniko te porowhita, e rua ōna tuaka ōrite: he tuaka matua (roa) me he tuaka iti (poto). Anei ētahi tauira o ngā raruraru me ngā porowhita.
Tauira Pātai 2
Hoatu he porowhita me te whārite \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). Whakatauhia te whārite o te rārangi pātapa i te pūwāhi \((2, \frac{3}{2})\) i runga i te porowhita.
Kōrero
Ko te whārite o te rārangi pātata ki te porowhita \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) i te pūwāhi \((x_1, y_1)\) ko:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Mā te whakamahi i te \(a = 4\) me te \(b = 3\), whakakapia ngā uara o \(a\), \(b\), me te pūwāhi \((2, \frac{3}{2})\):
\[ \frac{x(2)}{4^2} + \frac{y(\frac{3}{2})}{3^2} = 1 \]
\[ \frac{2x}{16} + \frac{3y}{6} = 1 \]
\[ \frac{x}{8} + \frac{y}{2} = 1 \]
Whakareatia te whārite katoa ki te 8 hei whakakore i ngā hautau:
\[ x + 4y = 8 \]
Nō reira, ko te whārite o te rārangi pātapa ki te porowhita ko \( x + 4y = 8 \).
Rārangi Pānga ki te Parabola
He wāhanga kōniko te parabola me te tuaka ōrite kotahi me te tihi kotahi. Anei ētahi tauira o ngā raruraru me ngā parabola.
Tauira Pātai 3
Hoatu he parabola me te whārite \( y^2 = 4x \). Whakatauhia te whārite o te rārangi pātapa i te pūwāhi \((1, 2)\) i runga i te parabola.
Kōrero
Ko te whārite o te rārangi pātata ki te parabola \( y^2 = 4ax \) i te pūwāhi \((x_1, y_1)\) ko:
\[ yy_1 = 2a(x + x_1) \]
Mai i te whārite parabola \( y^2 = 4x \), ka whiwhi tātou \( 4a = 4 \) kia \( a = 1 \). Whakakapia te uara o \( a \) me te pūwāhi \((1, 2)\):
\[ 2y = 2(1)(x + 1) \]
\[ 2y = 2x + 2 \]
\[ y = x + 1 \]
Nō reira, ko te whārite o te rārangi pātapa ki te parabola ko \( y = x + 1 \).
Rārangi Pānga ki te Pūwero
He wāhanga kōniko te hyperbola me ngā peka e rua me ngā asymptotes e rua. Anei ētahi tauira o ngā raruraru me ngā hyperbola.
Tauira Pātai 4
Hoatu he pūrua-pūrua me te whārite \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \). Whakatauhia te whārite o te rārangi pātapa i te pūwāhi \((5, 0)\) i runga i te pūrua-pūrua.
Kōrero
Ko te whārite o te rārangi pātata ki te hyperbola \(\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1\) i te pūwāhi \((x_1, y_1)\) ko:
\[ \frac{xx_1}{a^2} – \frac{yy_1}{b^2} = 1 \]
Mā te whakamahi i te \( a = 5 \) me te \( b = 4 \), whakakapia ngā uara \( a \), \( b \), me te pūwāhi \((5, 0)\):
\[ \frac{x(5)}{25} – \frac{y(0)}{16} = 1 \]
\[ \frac{5x}{25} – 0 = 1 \]
\[ \frac{x}{5} = 1 \]
\[ x = 5 \]
Nō reira, ko te whārite o te rārangi pātata ki te hyperbola ko \( x = 5 \).
Whakamutunga
He mea nui te mahi a ngā tapanga ki ngā wāhanga kōniko i roto i te pāngarau me ngā mahi whaihua. Ko te mārama ki te kimi i ngā whārite o ngā tapanga ki ngā momo wāhanga kōniko, pērā i ngā porowhita, ngā porowhita, ngā parabola, me ngā hyperbola, he pūkenga nui i roto i te tātaitai me te āhuahanga tātari. Mā ngā tauira me ngā kōrero i runga ake nei, ko te tumanako ka pai ake te mārama o ngā kaipānui ki ngā ariā me ngā tikanga mō te whakatau i ngā tapanga ki ngā wāhanga kōniko.
